Similar shapes

1. Similarity

ID is: 3071 Seed is: 1615

Identifying similar polygons

The diagram below shows two polygons. The figure is drawn to scale.

Are the shapes here similar to one another (select from the choices below)?
Select the correct reason which explains the first answer.


shape A	shape B

Answer:

Are the shapes similar?
The reason is:

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Similar objects must have the same shape, but might not be the same size.

STEP: Question 1: Compare the angles and sides of the two shapes
[−1 point ⇒ 1 / 2 points left]

The question shows us two shapes, and we need to decide if they are similar or not. In maths, two shapes are similar if corresponding angles are equal and if corresponding sides are all proportional (have the same ratio). In other words, if two objects are similar to each other, one of them can be "zoomed in or out" to make it identical to the other one.

These shapes are similar: the angles are all 90 degrees, and the sides are proportional because the sides of each shape are all equal: $\frac{9}{18} = \frac{9}{18}$ . In fact, any two equilateral squares must be similar.


shape A	shape B

These two shapes are similar. The correct choice from the list is: Yes.

STEP: Question 2: Choose the correct reason or reasons
[−1 point ⇒ 0 / 2 points left]

As noted in the above explanation, the shapes are similar. This can only be true if the corresponding angles are equal and the sides are all proportional: both must be true for similarity.

Therefore, the correct choice from the list is: Both of the above

Submit your answer as: and

ID is: 3071 Seed is: 7410

Identifying similar polygons

The diagram below shows two polygons. The figure is drawn to scale.

Are the shapes here similar to one another (select from the choices below)?
Select the correct reason which explains the first answer.


shape A	shape B

Answer:

Are the shapes similar?
The reason is:

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Similar objects must have the same shape, but might not be the same size.

STEP: Question 1: Compare the angles and sides of the two shapes
[−1 point ⇒ 1 / 2 points left]

The shapes here, which are kites, are not similar: corresponding angles are equal, but the sides are not proportional: $\frac{1}{2}$ is not equal to $\frac{3}{3}$ .


shape A	shape B

These two shapes are not similar. The correct choice from the list is: No.

STEP: Question 2: Choose the correct reason or reasons
[−1 point ⇒ 0 / 2 points left]

Therefore, the correct choice from the list is: The sides are not all proportional

Submit your answer as: and

ID is: 3071 Seed is: 2630

Identifying similar polygons

The diagram below shows two polygons. The figure is drawn to scale.

Determine whether the two polygons shown are similar or not.
What is the ratio of the size of polygon A to polygon B. Give your answer as a fraction. If the polygons are not similar, write 'None' for the ratio.


polygon A	polygon B

Answer:

Are the shapes similar?
The ratio of the lengths is:

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Similar objects must have the same shape, but might not be the same size.

STEP: Question 1: Compare the angles and sides of the two polygons
[−1 point ⇒ 1 / 2 points left]

These polygons, which are kites, are similar. We can see that the four angles are all equal, and the sides are all proportional: $\frac{1}{3} = \frac{3}{9}$ .


polygon A	polygon B

These two polygons are similar. The correct choice from the list is: Yes.

STEP: Question 2: Calculate the ratio of the shapes (if they are similar)
[−1 point ⇒ 0 / 2 points left]

Similar shapes always have proportional sides. We can use any pair of corresponding sides to find the ratio: $\frac{1}{3}$ .

The correct answer is: $\frac{1}{3}$ .

Submit your answer as: and

ID is: 4283 Seed is: 9494

Prove similarity in adjacent triangles

Consider the following diagram:

Prove that the triangles are similar.

Adebankole has already answered the question, and his proof is written below. But, Adebankole has made a mistake. Look carefully at his proof and identify where he has made his mistake.

Line	Proof
(a)	In $\Delta CAD$ and $\Delta ABD$ :
(b)	1. $\dfrac{CA}{AB} = \dfrac{\text{45}}{\text{20}} = \text{2.25}$
(c)	2. $\dfrac{DB}{DA} = \dfrac{\text{18}}{\text{8}} = \text{2.25}$
(d)	3. $\dfrac{CD}{AD} = \dfrac{\text{40.5}}{\text{18}} = \text{2.25}$
(e)	$\therefore \Delta CAD \;\|\|\|\; \Delta ABD$ (sides of $\Delta$ in prop)

Answer:

The mistake is on Line .

Replace this line with

1. $AD$ is common
1. $\dfrac{AB}{CA} = \dfrac{\text{45}}{\text{20}} = \text{2.25}$
2. $\dfrac{DA}{DB} = \dfrac{\text{18}}{\text{8}} = \text{2.25}$
$\therefore \Delta CAD \;|||\; \Delta ABD$ (SSS)

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Check each line of the proof carefully. Are the sides of the triangles being divided in a consistent order? Is the similarity stated correctly, with the correct reason?

STEP: Identify the error in the proof
[−1 point ⇒ 2 / 3 points left]

When we use sides of $\Delta$ in prop to prove that two triangles are similar, we must divide the sides in the same way each time.

Here, Adebankole divided the sides in $\Delta CAD$ by the sides in $\Delta ABD$ on Lines (b) and (d). But, on Line (c), Adebankole wrote the sides the other way around. So, the mistake is on Line (c).

NOTE: Adebankole substituted in the values for

\frac{D A}{D B}

$\dfrac{DA}{DB}$ , which is why he got the correct answer for the proportionality constant. But, this substitution did not match the labels he had written.

STEP: Correct the proof
[−2 points ⇒ 0 / 3 points left]

Adebankole should have divided the sides of $\Delta CAD$ by the sides of $\Delta ABD$ every time.

The correct proof is:

Line	Proof
(a)	In $\Delta CAD$ and $\Delta ABD$ :
(b)	1. $\dfrac{CA}{AB} = \dfrac{\text{45}}{\text{20}} = \text{2.25}$
(c)	2. $\color{ForestGreen}{\dfrac{DA}{DB} = \dfrac{\text{18}}{\text{8}} = \text{2.25}}$
(d)	3. $\dfrac{CD}{AD} = \dfrac{\text{40.5}}{\text{18}} = \text{2.25}$
(e)	$\therefore \Delta CAD \;\|\|\|\; \Delta ABD$ (sides of $\Delta$ in prop)

Submit your answer as: and

\frac{D A}{D B} = \frac{18}{8} = 2.25

$\dfrac{DA}{DB} = \dfrac{\text{18}}{\text{8}} = \text{2.25}$

ID is: 4283 Seed is: 1812

Prove similarity in adjacent triangles

Consider the following diagram:

Prove that the triangles are similar.

Chibuike has already answered the question, and his proof is written below. But, Chibuike has made a mistake. Look carefully at his proof and identify where he has made his mistake.

Line	Proof
(a)	In $\Delta PRS$ and $\Delta PQR$ :
(b)	1. $\dfrac{SP}{RP} = \dfrac{\text{13.5}}{\text{9}} = \text{1.5}$
(c)	2. $\dfrac{PR}{PQ} = \dfrac{\text{9}}{\text{6}} = \text{1.5}$
(d)	3. $\dfrac{RQ}{SR} = \dfrac{\text{7.5}}{\text{5}} = \text{1.5}$
(e)	$\therefore \Delta PRS \;\|\|\|\; \Delta PQR$ (sides of $\Delta$ in prop)

Answer:

The mistake is on Line .

Replace this line with

2. $\dfrac{PQ}{PR} = \dfrac{\text{9}}{\text{6}} = \text{1.5}$
$\therefore \Delta PRS \equiv \Delta PQR$ (sides of $\Delta$ in prop)
$\therefore \Delta PRS \;|||\; \Delta PQR$ (SSS)
3. $\dfrac{SR}{RQ} = \dfrac{\text{7.5}}{\text{5}} = \text{1.5}$

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Check each line of the proof carefully. Are the sides of the triangles being divided in a consistent order? Is the similarity stated correctly, with the correct reason?

STEP: Identify the error in the proof
[−1 point ⇒ 2 / 3 points left]

When we use sides of $\Delta$ in prop to prove that two triangles are similar, we must divide the sides in the same way each time.

Here, Chibuike divided the sides in $\Delta PRS$ by the sides in $\Delta PQR$ on Lines (b) and (c). But, on Line (d), Chibuike wrote the sides the other way around. So, the mistake is on Line (d).

NOTE: Chibuike substituted in the values for

\frac{S R}{R Q}

$\dfrac{SR}{RQ}$ , which is why he got the correct answer for the proportionality constant. But, this substitution did not match the labels he had written.

STEP: Correct the proof
[−2 points ⇒ 0 / 3 points left]

Chibuike should have divided the sides of $\Delta PRS$ by the sides of $\Delta PQR$ every time.

The correct proof is:

Line	Proof
(a)	In $\Delta PRS$ and $\Delta PQR$ :
(b)	1. $\dfrac{SP}{RP} = \dfrac{\text{13.5}}{\text{9}} = \text{1.5}$
(c)	2. $\dfrac{PR}{PQ} = \dfrac{\text{9}}{\text{6}} = \text{1.5}$
(d)	3. $\color{ForestGreen}{\dfrac{SR}{RQ} = \dfrac{\text{7.5}}{\text{5}} = \text{1.5}}$
(e)	$\therefore \Delta PRS \;\|\|\|\; \Delta PQR$ (sides of $\Delta$ in prop)

Submit your answer as: and

\frac{S R}{R Q} = \frac{7.5}{5} = 1.5

$\dfrac{SR}{RQ} = \dfrac{\text{7.5}}{\text{5}} = \text{1.5}$

ID is: 4283 Seed is: 6871

Prove similarity in adjacent triangles

Consider the following diagram:

Prove that the triangles are similar.

Chichi has already answered the question, and her proof is written below. But, Chichi has made a mistake. Look carefully at her proof and identify where she has made her mistake.

Line	Proof
(a)	In $\Delta LJK$ and $\Delta MLK$ :
(b)	1. $\dfrac{JK}{LK} = \dfrac{\text{87.5}}{\text{35}} = \text{2.5}$
(c)	2. $\dfrac{KL}{KM} = \dfrac{\text{35}}{\text{14}} = \text{2.5}$
(d)	3. $\dfrac{JL}{LM} = \dfrac{\text{90}}{\text{36}} = \text{2.5}$
(e)	$\therefore \Delta LJK \equiv \Delta MLK$ (sides of $\Delta$ in prop)

Answer:

The mistake is on Line .

Replace this line with

$\therefore \Delta LJK \;|||\; \Delta MLK$ (equiangular $\Delta$ s)
1. $\dfrac{LK}{JK} = \dfrac{\text{87.5}}{\text{35}} = \text{2.5}$
$\therefore \Delta LJK \equiv \Delta MLK$ (SSS)
$\therefore \Delta LJK \;|||\; \Delta MLK$ (sides of $\Delta$ in prop)

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Check each line of the proof carefully. Are the sides of the triangles being divided in a consistent order? Is the similarity stated correctly, with the correct reason?

STEP: Identify the error in the proof
[−1 point ⇒ 2 / 3 points left]

Chichi has proved that the triangles are similar because their sides are in proportion. But, on Line (e), Chichi has used the symbol for congruency: $\equiv$ . So, the mistake is on Line (e).

TIP: The symbol for congruency (

\equiv

$\equiv$ ) looks like the equals sign (

=

$=$ ) but with an extra line. We can think of congruent triangles as being equal in every way. So, you can remember the symbol for congruency because it is almost like the equals sign.

STEP: Correct the proof
[−2 points ⇒ 0 / 3 points left]

Chichi should have used the symbol for similar triangles: $|||$ .

The correct proof is:

Line	Proof
(a)	In $\Delta LJK$ and $\Delta MLK$ :
(b)	1. $\dfrac{JK}{LK} = \dfrac{\text{87.5}}{\text{35}} = \text{2.5}$
(c)	2. $\dfrac{KL}{KM} = \dfrac{\text{35}}{\text{14}} = \text{2.5}$
(d)	3. $\dfrac{JL}{LM} = \dfrac{\text{90}}{\text{36}} = \text{2.5}$
(e)	$\color{ForestGreen}{\therefore \Delta LJK \;\|\|\|\; \Delta MLK}$ (sides of $\color{ForestGreen}{\Delta}$ in prop)

Submit your answer as: and

∴ Δ L J K | | | Δ M L K

$\therefore \Delta LJK \;|||\; \Delta MLK$ (sides of

Δ

$\Delta$ in prop)

ID is: 4289 Seed is: 6590

Consequences of similarity

In the diagram below, $\Delta JKL \;|||\; \Delta PMN$ .

Determine the length of $PN$ .

Answer:

P N =

$PN =$

units

$\text{units}$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

If two triangles are similar, then

they are equiangular (their matching angles are equal), and
their sides are in proportion (you always have to multiply by the same number to get the matching side in the other triangle).

STEP: Match up the sides and angles
[−2 points ⇒ 0 / 2 points left]

If two triangles are similar then

they are equiangular (their matching angles are equal), and
their sides are in proportion (you always have to multiply by the same number to get the matching side in the other triangle).

We know that the triangles are similar, so their sides will be in proportion. We use this fact to determine the length of $PN$ .

First we match up the sides. In the statement $\Delta JKL \;|||\; \Delta PMN$ , ${\color{red}{JK}}$ and ${\color{red}{PM}}$ are the first two letters, so their sides must match up. Similarly, ${\color{blue}{KL}}$ and ${\color{blue}{MN}}$ match up because their letters came last. Finally, ${\color{ForestGreen}{JL}}$ and ${\color{ForestGreen}{PN}}$ must match up, because they are the only sides left.

Since $JK$ and $PM$ match up, we can work out the proportionality constant:

\begin{aligned} k & = \frac{P M}{J K} \\ ∴ k & = \frac{49}{28} \\ ∴ k & = 1.75 \end{aligned}

$\begin{aligned} k & = \dfrac{PM}{JK} \\ \therefore k & = \dfrac{49}{28} \\ \therefore k & = \text{1.75} \end{aligned}$

This is the number by which all sides in $\Delta JKL$ should by multiplied in order to get the matching sides in $\Delta PMN$ , because $28 \times \text{1.75} = 49$ .

This is only true because we were told that $\Delta JKL \;|||\; \Delta PMN$ . So, we use this as our reason.

So,

\begin{aligned} P N & = 44 \times 1.75 \\ ∴ P N & = 77 units (Δ J K L | | | Δ P M N) \end{aligned}

$\begin{aligned} PN & = 44 \times \text{1.75} \\ \therefore PN & = \text{77}\ \text{units} \;\; (\Delta JKL \;|||\; \Delta PMN) \end{aligned}$

Submit your answer as: and

ID is: 4289 Seed is: 2989

Consequences of similarity

In the diagram below, $\Delta XYZ \;|||\; \Delta LJK$ .

Determine the length of $LK$ .

Answer:

L K =

$LK =$

units

$\text{units}$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

If two triangles are similar, then

they are equiangular (their matching angles are equal), and
their sides are in proportion (you always have to multiply by the same number to get the matching side in the other triangle).

STEP: Match up the sides and angles
[−2 points ⇒ 0 / 2 points left]

If two triangles are similar then

they are equiangular (their matching angles are equal), and
their sides are in proportion (you always have to multiply by the same number to get the matching side in the other triangle).

We know that the triangles are similar, so their sides will be in proportion. We use this fact to determine the length of $LK$ .

First we match up the sides. In the statement $\Delta XYZ \;|||\; \Delta LJK$ , ${\color{red}{XY}}$ and ${\color{red}{LJ}}$ are the first two letters, so their sides must match up. Similarly, ${\color{blue}{YZ}}$ and ${\color{blue}{JK}}$ match up because their letters came last. Finally, ${\color{ForestGreen}{XZ}}$ and ${\color{ForestGreen}{LK}}$ must match up, because they are the only sides left.

Since $XY$ and $LJ$ match up, we can work out the proportionality constant:

\begin{aligned} k & = \frac{L J}{X Y} \\ ∴ k & = \frac{45}{18} \\ ∴ k & = 2.5 \end{aligned}

$\begin{aligned} k & = \dfrac{LJ}{XY} \\ \therefore k & = \dfrac{45}{18} \\ \therefore k & = \text{2.5} \end{aligned}$

This is the number by which all sides in $\Delta XYZ$ should by multiplied in order to get the matching sides in $\Delta LJK$ , because $18 \times \text{2.5} = 45$ .

This is only true because we were told that $\Delta XYZ \;|||\; \Delta LJK$ . So, we use this as our reason.

So,

\begin{aligned} L K & = 20 \times 2.5 \\ ∴ L K & = 50 units (Δ X Y Z | | | Δ L J K) \end{aligned}

$\begin{aligned} LK & = 20 \times \text{2.5} \\ \therefore LK & = \text{50}\ \text{units} \;\; (\Delta XYZ \;|||\; \Delta LJK) \end{aligned}$

Submit your answer as: and

ID is: 4289 Seed is: 2990

Consequences of similarity

In the diagram below, $\Delta ABC \;|||\; \Delta NMP$ .

Determine the length of $NP$ .

Answer:

N P =

$NP =$

units

$\text{units}$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

If two triangles are similar, then

they are equiangular (their matching angles are equal), and
their sides are in proportion (you always have to multiply by the same number to get the matching side in the other triangle).

STEP: Match up the sides and angles
[−2 points ⇒ 0 / 2 points left]

If two triangles are similar then

they are equiangular (their matching angles are equal), and
their sides are in proportion (you always have to multiply by the same number to get the matching side in the other triangle).

We know that the triangles are similar, so their sides will be in proportion. We use this fact to determine the length of $NP$ .

First we match up the sides. In the statement $\Delta ABC \;|||\; \Delta NMP$ , ${\color{red}{AB}}$ and ${\color{red}{NM}}$ are the first two letters, so their sides must match up. Similarly, ${\color{blue}{BC}}$ and ${\color{blue}{MP}}$ match up because their letters came last. Finally, ${\color{ForestGreen}{AC}}$ and ${\color{ForestGreen}{NP}}$ must match up, because they are the only sides left.

Since $AB$ and $NM$ match up, we can work out the proportionality constant:

\begin{aligned} k & = \frac{N M}{A B} \\ ∴ k & = \frac{90}{40} \\ ∴ k & = 2.25 \end{aligned}

$\begin{aligned} k & = \dfrac{NM}{AB} \\ \therefore k & = \dfrac{90}{40} \\ \therefore k & = \text{2.25} \end{aligned}$

This is the number by which all sides in $\Delta ABC$ should by multiplied in order to get the matching sides in $\Delta NMP$ , because $40 \times \text{2.25} = 90$ .

This is only true because we were told that $\Delta ABC \;|||\; \Delta NMP$ . So, we use this as our reason.

So,

\begin{aligned} N P & = 32 \times 2.25 \\ ∴ N P & = 72 units (Δ A B C | | | Δ N M P) \end{aligned}

$\begin{aligned} NP & = 32 \times \text{2.25} \\ \therefore NP & = \text{72}\ \text{units} \;\; (\Delta ABC \;|||\; \Delta NMP) \end{aligned}$

Submit your answer as: and

ID is: 4302 Seed is: 7041

Identify similar triangles

In the diagram below, $ST \bot TR$ and $SR \bot QP$ . Also, $\hat{R} = x$ , $RQ = 7$ , $QT = 4$ , and $PR = 6$ .

Ekene needs to prove that $\Delta STR \;|||\; \Delta QPR$ . He has already started, and his incomplete proof is written below:

In $\Delta STR$ and $\Delta QPR$ :
1. $\hat{R}$ is common
2. $S\hat{T}R = Q\hat{P}R = 90°$ (given)
  ...
How should Ekene complete the proof? Choose the best option.
Answer:
3. $T\hat{Q}P = 90° + x$ (ext $\angle$ of $\Delta$ )

$\therefore \Delta STR \;|||\; \Delta QPR$ (equiangular $\Delta$ s)

3. $\hat{S} = 180° - x$ (sum of $\angle$ s in $\Delta$ )
Also, $R\hat{Q}P = 180° - x$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{S} = R\hat{Q}P$ (both equal to $180° - x$ )

$\therefore \Delta STR \;|||\; \Delta QPR$ (equiangular $\Delta$ s)

3. $\hat{S} = 90° - x$ (sum of $\angle$ s in $\Delta$ )
Also, $R\hat{Q}P = 90° - x$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{S} = R\hat{Q}P$ (both equal to $90° - x$ )

$\therefore \Delta STR \;|||\; \Delta QPR$ (equiangular $\Delta$ s)

3. $\dfrac{TR}{PR} = \dfrac{11}{6}$

$\therefore \Delta STR \;|||\; \Delta QPR$ (sides of $\Delta$ in prop)
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

There are two ways of proving similarity in triangles:
- Prove that the two triangles are equiangular.
- Prove that the two triangles have matching sides in proportion.
Which can you use here?

STEP: Choose the correct proof
[−3 points ⇒ 0 / 3 points left]
NOTE:
There are two ways of proving similarity in triangles:
- Prove that the two triangles are equiangular.
- Prove that the two triangles have matching sides in proportion.
Ekene has already proved two pairs of angles equal for us, and we need to prove the third pair. To do so, it is helpful to look at each triangle separately.

Looking at $\Delta STR$ , we can write $\hat{S}$ in terms of $x$ :

$\begin{aligned} {\color{ForestGreen}{\hat{S}}} + {\color{blue}{x}} + {\color{red}{90°}} & = 180° \;\; (\text{sum of } \angle \text{s in } \Delta) \\ \therefore \hat{S} + x + 90° - 90° & = 18 0° - 90° \\ \therefore \hat{S} + x & = 90° \\ \therefore \hat{S} + x - x & = 90° - x \\ \therefore \hat{S} & = 90° - x \end{aligned}$

In the same way in $\Delta QPR$ , we can write $R\hat{Q}P$ in terms of $x$ .

$\begin{aligned} {\color{ForestGreen}{R\hat{Q}P}} + {\color{blue}{x}} + {\color{red}{90°}} & = 180° \;\; (\text{sum of } \angle \text{s in } \Delta) \\ \therefore R\hat{Q}P & = 90° - x \end{aligned}$

NOTE: It is true that $T\hat{Q}P$ is equal to $90° + x$ . But, this is not an angle in either triangle, so it does not help us to prove that $\Delta STR \;|||\; \Delta QPR$ .

The full proof is shown below:

In $\color{gray}{\Delta STR}$ and $\color{gray}{\Delta QPR}$ :
1. $\color{gray}{\hat{R}}$ is common
2. $\color{gray}{S\hat{T}R = Q\hat{P}R = 90°}$ (given)
3. $\hat{S} = 90° - x$ (sum of $\angle$ s in $\Delta$ )
  Also, $R\hat{Q}P = 90° - x$ (sum of $\angle$ s in $\Delta$ )
  $\therefore \hat{S} = R\hat{Q}P$ (both equal to $90° - x$ )
$\therefore \Delta STR \;|||\; \Delta QPR$ (equiangular $\Delta$ s)

Submit your answer as:

3. $\hat{S} = 90° - x$ (sum of $\angle$ s in $\Delta$ )
Also, $R\hat{Q}P = 90° - x$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{S} = R\hat{Q}P$ (both equal to $90° - x$ )

$\therefore \Delta STR \;|||\; \Delta QPR$ (equiangular $\Delta$ s)
Hence, determine the length of $SP$ .

The diagram is repeated here for your convenience.

INSTRUCTION: Give your answer as a fraction, where necessary.

Answer: $SP =$
one-of
type(numeric.abserror(0.01))
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Since the triangles are similar, you know that the sides are in proportion. Use this fact to create an equation which connects $SP$ to the sides that you already know. Use the similarity statement to choose the correct pair of ratios.

STEP: Use the similarity statement to choose the correct pair of ratios
[−2 points ⇒ 2 / 4 points left]

We are trying to find $SP$ . Even though $SP$ is not a side in the similar triangles, it is part of side $SR$ :

$\begin{array}{c} SR = SP + PR \end{array}$

To make the algebra easier, let $SP = y$ .

Now we have

$\begin{array}{c} SR = y + 6 \end{array}$

The similarity is easier to see if we separate the triangles.

We can see from the similarity statement that $\color{blue}{SR}$ matches with $\color{blue}{QR}$ : $\Delta {\color{blue}{S}}T{\color{blue}{R}} \;|||\; \Delta {\color{blue}{Q}}P{\color{blue}{R}}$ .

We were given the lengths of $RQ$ and $QT$ . Since $TR = RQ + QT$ , we know that $TR = 11$ . So we know the length of ${\color{ForestGreen}{TR}}$ and ${\color{ForestGreen}{PR}}$ , which match up: $\Delta S{\color{ForestGreen}{T}}{\color{ForestGreen}{R}} \;|||\; \Delta Q{\color{ForestGreen}{P}}{\color{ForestGreen}{R}}$

So, dividing the sides of the big triangle by the sides of the small triangle, we get:

$\begin{aligned} \dfrac{SR}{QR} & = \dfrac{TR}{PR} \;\; (\Delta STR \;|||\; \Delta QPR) \end{aligned}$

NOTE: We only know that the fractions are equal because the triangles are similar. So, we write this as the reason.

STEP: Solve the equation using inverse operations
[−2 points ⇒ 0 / 4 points left]

First we need to substitute in the values from the diagram. As long as there are only numbers and $y$ -terms, we will be able to solve the equation.

$\begin{aligned} {\color{blue}{\dfrac{y + 6}{7}}} & = {\color{ForestGreen}{\dfrac{11}{6}}} \\ \dfrac{y + 6}{7} \times 7 & = \dfrac{11}{6} \times 7 \\ y + 6 & = \dfrac{77}{6} \\ y + 6 - 6 & = \dfrac{77}{6} - 6 \\ \therefore y & = \frac{41}{6} \end{aligned}$

Submit your answer as: and

ID is: 4302 Seed is: 6366

Identify similar triangles

In the diagram below, $RS \bot ST$ and $RT \bot SQ$ . Also, $\hat{T} = x$ , $ST = 8$ , and $QT = 4$ .

Chibueze needs to prove that $\Delta RST \;|||\; \Delta SQT$ . He has already started, and his incomplete proof is written below:

In $\Delta RST$ and $\Delta SQT$ :
1. $\hat{T}$ is common
2. $R\hat{S}T = S\hat{Q}T = 90°$ (given)
  ...
How should Chibueze complete the proof? Choose the best option.
Answer:
3. $\hat{R} = 90° - x$ (sum of $\angle$ s in $\Delta$ )
Also, $T\hat{S}Q = 90° - x$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{R} = T\hat{S}Q$ (both equal to $90° - x$ )

$\therefore \Delta RST \;|||\; \Delta SQT$ (equiangular $\Delta$ s)

3. $\hat{R} = 180° - x$ (sum of $\angle$ s in $\Delta$ )
Also, $T\hat{S}Q = 180° - x$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{R} = T\hat{S}Q$ (both equal to $180° - x$ )

$\therefore \Delta RST \;|||\; \Delta SQT$ (equiangular $\Delta$ s)

3. $\dfrac{ST}{QT} = \dfrac{8}{4}$

$\therefore \Delta RST \;|||\; \Delta SQT$ (sides of $\Delta$ in prop)

3. $T\hat{S}Q = 45°$ (given)
$\therefore \hat{R} = 45°$ ( $\Delta RST \;|||\; \Delta SQT$ )
$\therefore \hat{R} = R\hat{S}Q$ (both equal to $45°$ )

$\therefore \Delta RST \;|||\; \Delta SQT$ (equiangular $\Delta$ s)
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

There are two ways of proving similarity in triangles:
- Prove that the two triangles are equiangular.
- Prove that the two triangles have matching sides in proportion.
Which can you use here?

STEP: Choose the correct proof
[−3 points ⇒ 0 / 3 points left]
NOTE:
There are two ways of proving similarity in triangles:
- Prove that the two triangles are equiangular.
- Prove that the two triangles have matching sides in proportion.
Chibueze has already proved two pairs of angles equal for us, and we need to prove the third pair. To do so, it is helpful to look at each triangle separately.

Looking at $\Delta RST$ , we can write $\hat{R}$ in terms of $x$ :

$\begin{aligned} {\color{ForestGreen}{\hat{R}}} + {\color{blue}{x}} + {\color{red}{90°}} & = 180° \;\; (\text{sum of } \angle \text{s in } \Delta) \\ \therefore \hat{R} + x + 90° - 90° & = 18 0° - 90° \\ \therefore \hat{R} + x & = 90° \\ \therefore \hat{R} + x - x & = 90° - x \\ \therefore \hat{R} & = 90° - x \end{aligned}$

In the same way in $\Delta SQT$ , we can write $T\hat{S}Q$ in terms of $x$ .

$\begin{aligned} {\color{ForestGreen}{T\hat{S}Q}} + {\color{blue}{x}} + {\color{red}{90°}} & = 180° \;\; (\text{sum of } \angle \text{s in } \Delta) \\ \therefore T\hat{S}Q & = 90° - x \end{aligned}$

NOTE: We cannot assume that $QS$ cuts $R\hat{S}T$ in half. $T\hat{S}Q$ might be equal to $45°$ , or it might be something else. We must only use the information that we have been given.

The full proof is shown below:

In $\color{gray}{\Delta RST}$ and $\color{gray}{\Delta SQT}$ :
1. $\color{gray}{\hat{T}}$ is common
2. $\color{gray}{R\hat{S}T = S\hat{Q}T = 90°}$ (given)
3. $\hat{R} = 90° - x$ (sum of $\angle$ s in $\Delta$ )
  Also, $T\hat{S}Q = 90° - x$ (sum of $\angle$ s in $\Delta$ )
  $\therefore \hat{R} = T\hat{S}Q$ (both equal to $90° - x$ )
$\therefore \Delta RST \;|||\; \Delta SQT$ (equiangular $\Delta$ s)

Submit your answer as:

3. $\hat{R} = 90° - x$ (sum of $\angle$ s in $\Delta$ )
Also, $T\hat{S}Q = 90° - x$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{R} = T\hat{S}Q$ (both equal to $90° - x$ )

$\therefore \Delta RST \;|||\; \Delta SQT$ (equiangular $\Delta$ s)
Hence, determine the length of $RQ$ .

The diagram is repeated here for your convenience.

INSTRUCTION: Give your answer as a fraction, where necessary.

Answer: $RQ =$
one-of
type(numeric.abserror(0.01))
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Since the triangles are similar, you know that the sides are in proportion. Use this fact to create an equation which connects $RQ$ to the sides that you already know. Use the similarity statement to choose the correct pair of ratios.

STEP: Use the similarity statement to choose the correct pair of ratios
[−2 points ⇒ 2 / 4 points left]

We are trying to find $RQ$ . Even though $RQ$ is not a side in the similar triangles, it is part of side $RT$ :

$\begin{array}{c} RT = RQ + QT \end{array}$

To make the algebra easier, let $RQ = y$ .

Now we have

$\begin{array}{c} RT = y + 4 \end{array}$

The similarity is easier to see if we separate the triangles.

We can see from the similarity statement that $\color{blue}{RT}$ matches with $\color{blue}{ST}$ : $\Delta {\color{blue}{R}}S{\color{blue}{T}} \;|||\; \Delta {\color{blue}{S}}Q{\color{blue}{T}}$ .

We know the lengths of ${\color{ForestGreen}{ST}}$ and ${\color{ForestGreen}{QT}}$ . They are a matching pair of sides: ( $\Delta R{\color{ForestGreen}{S}}{\color{ForestGreen}{T}} \;|||\; \Delta S{\color{ForestGreen}{Q}}{\color{ForestGreen}{T}}$ ).

So, dividing the sides of the big triangle by the sides of the small triangle, we get:

$\begin{aligned} \dfrac{RT}{ST} & = \dfrac{ST}{QT} \;\; (\Delta RST \;|||\; \Delta SQT) \end{aligned}$

NOTE: We only know that the fractions are equal because the triangles are similar. So, we write this as the reason.

STEP: Solve the equation using inverse operations
[−2 points ⇒ 0 / 4 points left]

First we need to substitute in the values from the diagram. As long as there are only numbers and $y$ -terms, we will be able to solve the equation.

$\begin{aligned} {\color{blue}{\dfrac{y + 4}{8}}} & = {\color{ForestGreen}{\dfrac{8}{4}}} \\ \dfrac{y + 4}{8} \times 8 & = \dfrac{8}{4} \times 8 \\ y + 4 & = \dfrac{64}{4} \\ y + 4 - 4 & = \dfrac{64}{4} - 4 \\ \therefore y & = 12 \end{aligned}$

Submit your answer as: and

ID is: 4302 Seed is: 4622

Identify similar triangles

In the diagram below, $TS \bot SQ$ and $TQ \bot PR$ . Also, $\hat{Q} = x$ , $QP = 6$ , $PS = 4$ , and $RQ = 5$ .

Justine needs to prove that $\Delta TSQ \;|||\; \Delta PRQ$ . She has already started, and her incomplete proof is written below:

In $\Delta TSQ$ and $\Delta PRQ$ :
1. $\hat{Q}$ is common
2. $T\hat{S}Q = P\hat{R}Q = 90°$ (given)
  ...
How should Justine complete the proof? Choose the best option.
Answer:
3. $\hat{T} = 90° - x$ (sum of $\angle$ s in $\Delta$ )
Also, $Q\hat{P}R = 90° - x$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{T} = Q\hat{P}R$ (both equal to $90° - x$ )

$\therefore \Delta TSQ \;|||\; \Delta PRQ$ (equiangular $\Delta$ s)

3. $\dfrac{SQ}{RQ} = \dfrac{10}{5}$

$\therefore \Delta TSQ \;|||\; \Delta PRQ$ (sides of $\Delta$ in prop)

3. $S\hat{P}R = 90° + x$ (ext $\angle$ of $\Delta$ )

$\therefore \Delta TSQ \;|||\; \Delta PRQ$ (equiangular $\Delta$ s)

3. $\hat{T} = 180° - x$ (sum of $\angle$ s in $\Delta$ )
Also, $Q\hat{P}R = 180° - x$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{T} = Q\hat{P}R$ (both equal to $180° - x$ )

$\therefore \Delta TSQ \;|||\; \Delta PRQ$ (equiangular $\Delta$ s)
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

There are two ways of proving similarity in triangles:
- Prove that the two triangles are equiangular.
- Prove that the two triangles have matching sides in proportion.
Which can you use here?

STEP: Choose the correct proof
[−3 points ⇒ 0 / 3 points left]
NOTE:
There are two ways of proving similarity in triangles:
- Prove that the two triangles are equiangular.
- Prove that the two triangles have matching sides in proportion.
Justine has already proved two pairs of angles equal for us, and we need to prove the third pair. To do so, it is helpful to look at each triangle separately.

Looking at $\Delta TSQ$ , we can write $\hat{T}$ in terms of $x$ :

$\begin{aligned} {\color{ForestGreen}{\hat{T}}} + {\color{blue}{x}} + {\color{red}{90°}} & = 180° \;\; (\text{sum of } \angle \text{s in } \Delta) \\ \therefore \hat{T} + x + 90° - 90° & = 18 0° - 90° \\ \therefore \hat{T} + x & = 90° \\ \therefore \hat{T} + x - x & = 90° - x \\ \therefore \hat{T} & = 90° - x \end{aligned}$

In the same way in $\Delta PRQ$ , we can write $Q\hat{P}R$ in terms of $x$ .

$\begin{aligned} {\color{ForestGreen}{Q\hat{P}R}} + {\color{blue}{x}} + {\color{red}{90°}} & = 180° \;\; (\text{sum of } \angle \text{s in } \Delta) \\ \therefore Q\hat{P}R & = 90° - x \end{aligned}$

NOTE: It is true that $S\hat{P}R$ is equal to $90° + x$ . But, this is not an angle in either triangle, so it does not help us to prove that $\Delta TSQ \;|||\; \Delta PRQ$ .

The full proof is shown below:

In $\color{gray}{\Delta TSQ}$ and $\color{gray}{\Delta PRQ}$ :
1. $\color{gray}{\hat{Q}}$ is common
2. $\color{gray}{T\hat{S}Q = P\hat{R}Q = 90°}$ (given)
3. $\hat{T} = 90° - x$ (sum of $\angle$ s in $\Delta$ )
  Also, $Q\hat{P}R = 90° - x$ (sum of $\angle$ s in $\Delta$ )
  $\therefore \hat{T} = Q\hat{P}R$ (both equal to $90° - x$ )
$\therefore \Delta TSQ \;|||\; \Delta PRQ$ (equiangular $\Delta$ s)

Submit your answer as:

3. $\hat{T} = 90° - x$ (sum of $\angle$ s in $\Delta$ )
Also, $Q\hat{P}R = 90° - x$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{T} = Q\hat{P}R$ (both equal to $90° - x$ )

$\therefore \Delta TSQ \;|||\; \Delta PRQ$ (equiangular $\Delta$ s)
Hence, determine the length of $TR$ .

The diagram is repeated here for your convenience.

INSTRUCTION: Give your answer as a fraction, where necessary.

Answer: $TR =$
one-of
type(numeric.abserror(0.01))
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Since the triangles are similar, you know that the sides are in proportion. Use this fact to create an equation which connects $TR$ to the sides that you already know. Use the similarity statement to choose the correct pair of ratios.

STEP: Use the similarity statement to choose the correct pair of ratios
[−2 points ⇒ 2 / 4 points left]

We are trying to find $TR$ . Even though $TR$ is not a side in the similar triangles, it is part of side $TQ$ :

$\begin{array}{c} TQ = TR + RQ \end{array}$

To make the algebra easier, let $TR = y$ .

Now we have

$\begin{array}{c} TQ = y + 5 \end{array}$

The similarity is easier to see if we separate the triangles.

We can see from the similarity statement that $\color{blue}{TQ}$ matches with $\color{blue}{PQ}$ : $\Delta {\color{blue}{T}}S{\color{blue}{Q}} \;|||\; \Delta {\color{blue}{P}}R{\color{blue}{Q}}$ .

We were given the lengths of $QP$ and $PS$ . Since $SQ = QP + PS$ , we know that $SQ = 10$ . So we know the length of ${\color{ForestGreen}{SQ}}$ and ${\color{ForestGreen}{RQ}}$ , which match up: $\Delta T{\color{ForestGreen}{S}}{\color{ForestGreen}{Q}} \;|||\; \Delta P{\color{ForestGreen}{R}}{\color{ForestGreen}{Q}}$

So, dividing the sides of the big triangle by the sides of the small triangle, we get:

$\begin{aligned} \dfrac{TQ}{PQ} & = \dfrac{SQ}{RQ} \;\; (\Delta TSQ \;|||\; \Delta PRQ) \end{aligned}$

NOTE: We only know that the fractions are equal because the triangles are similar. So, we write this as the reason.

STEP: Solve the equation using inverse operations
[−2 points ⇒ 0 / 4 points left]

First we need to substitute in the values from the diagram. As long as there are only numbers and $y$ -terms, we will be able to solve the equation.

$\begin{aligned} {\color{blue}{\dfrac{y + 5}{6}}} & = {\color{ForestGreen}{\dfrac{10}{5}}} \\ \dfrac{y + 5}{6} \times 6 & = \dfrac{10}{5} \times 6 \\ y + 5 & = \dfrac{60}{5} \\ y + 5 - 5 & = \dfrac{60}{5} - 5 \\ \therefore y & = 7 \end{aligned}$

Submit your answer as: and

ID is: 4258 Seed is: 3184

Definition: Similarity

Two triangles are similar if:

Answer:

some of their angles are equal.
some of their sides are the same length.
they are exactly the same shape, but they may be different sizes.
their angles are in proportion.

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

What do we mean in maths when we say that two triangles are similar?

STEP: Identify the definition of similarity
[−2 points ⇒ 0 / 2 points left]

We say that two triangles are similar if they are exactly the same shape, but one might be bigger or smaller than the other one. Another way of saying this is that one is an enlargement of the other one.

For example, $\Delta PQR$ and $\Delta XYZ$ below are the same shape. But, they are different sizes.

The sides of similar triangles are in proportion. That means that all of the sides in one triangle have been multiplied by the same number to get the sides in the second triangle. All of the sides in $\Delta PQR$ have been multiplied by $\text{2}$ to get $\Delta XYZ$ .

Similar triangles are the same shape, so they always have the same angles. We can see the matching angles in $\Delta PQR$ and $\Delta XYZ$ .

NOTE: In normal English, we say two things are similar if they are almost the same. But, in Maths, it is not precise enough to say that similar shapes are approximately the same. In Maths, similarity has a very specific meaning.

Two triangles are similar if either

their matching sides are in proportion, or
their matching angles are equal.

NOTE: In many polygons, we have to prove that the matching sides are in proportion and the matching angles are equal, if we want to be sure that the shapes are similar. But, in triangles, we only need to prove that one of them is true. This is a special property of triangles, which you will learn more about in grade

12

$\text{12}$ .

Submit your answer as:

they are exactly the same shape, but they may be different sizes.

ID is: 4258 Seed is: 7198

Definition: Similarity

Two triangles are similar if:

Answer:

they are approximately the same shape.
they are exactly the same shape, but they may be different sizes.
some of their sides are the same length.
some of their angles are equal.

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

What do we mean in maths when we say that two triangles are similar?

STEP: Identify the definition of similarity
[−2 points ⇒ 0 / 2 points left]

For example, $\Delta PQR$ and $\Delta XYZ$ below are the same shape. But, they are different sizes.

Similar triangles are the same shape, so they always have the same angles. We can see the matching angles in $\Delta PQR$ and $\Delta XYZ$ .

Two triangles are similar if either

their matching sides are in proportion, or
their matching angles are equal.

12

$\text{12}$ .

Submit your answer as:

they are exactly the same shape, but they may be different sizes.

ID is: 4258 Seed is: 3679

Definition: Similarity

Two triangles are similar if:

Answer:

their angles are in proportion.
their matching angles are equal.
some of their sides are the same length.
some of their angles are equal.

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

What do we mean in maths when we say that two triangles are similar?

STEP: Identify the definition of similarity
[−2 points ⇒ 0 / 2 points left]

For example, $\Delta ABC$ and $\Delta DEF$ below are the same shape. But, they are different sizes.

The sides of similar triangles are in proportion. That means that all of the sides in one triangle have been multiplied by the same number to get the sides in the second triangle. All of the sides in $\Delta ABC$ have been multiplied by $\text{2}$ to get $\Delta DEF$ .

Similar triangles are the same shape, so they always have the same angles. We can see the matching angles in $\Delta ABC$ and $\Delta DEF$ .

Two triangles are similar if either

their matching sides are in proportion, or
their matching angles are equal.

12

$\text{12}$ .

Submit your answer as:

their matching angles are equal.

ID is: 3091 Seed is: 7688

Using the proportionality of similar polygons

In the diagram below, $JKNP\; ||| \;$ $LMPQ$ . Compute the value of $f$ . The diagram is not necessarily drawn to scale.

Answer:

f =

$f =$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Similar shapes have proportional lengths. Write a proportion for the sides of the polygons and then solve the equation for $f$ .

STEP: Identify corresponding sides and write a proportion for these sides
[−1 point ⇒ 2 / 3 points left]

We have a diagram with quadrilaterals, and we need to figure out the value of the variable $f$ . The question tells us that the quadrilaterals are similar, which means that the sides of the shapes are proportional. Therefore, we can write a proportion for the sides of the polygons and solve it for $f$ .

There are a number of different proportions that we can write to summarise the lengths of the sides in the diagram: we can write a proportion for any pairs of corresponding sides. Here are two proportions which are accurate (but there are others):

\begin{matrix} \frac{J K}{L Q} = \frac{N P}{M P} \frac{J K}{N P} = \frac{L Q}{M P} \end{matrix}

$\begin{array}{c}\frac{JK}{LQ} = \frac{NP}{MP} \qquad \frac{JK}{NP} = \frac{LQ}{MP}\end{array}$

STEP: Solve the equation for

f

$f$
[−2 points ⇒ 0 / 3 points left]

We will use the first proportion from above to find the value of $f$ . If you use a different correct proportion and solve it without any mistakes, you will get the correct answer for $f$ (and full marks).

Remember:

\begin{aligned} M P & = N P + M N \\ = - f + 3 + 2 \\ = - f + 5 \end{aligned}

$\begin{aligned} MP &= NP + MN \\ &= - f + 3 + 2 \\ &= - f + 5 \end{aligned}$

So now we can calculate $f$ :

\begin{aligned} \frac{J K}{L Q} & = \frac{N P}{M P} \\ \frac{2 f}{2 f + 4} & = \frac{- f + 3}{- f + 5} \\ (2 f) (- f + 5) & = (- f + 3) (2 f + 4) \\ - 2 f^{2} + 10 f & = - 2 f^{2} + 2 f + 12 ⟵_{cancel}^{- 2 f^{2} terms} \\ 8 f & = 12 \\ f & = \frac{12}{8} \\ f & = \frac{3}{2} \end{aligned}

$\begin{aligned} \frac{JK}{LQ} & = \frac{NP}{MP} \\ \frac{2 f}{2 f + 4} & = \frac{- f + 3}{- f + 5} \\ \left( 2 f \right) \left( - f + 5 \right) & = \left(- f + 3 \right) \left( 2 f + 4 \right) \\ - 2 f^{2} + 10 f & = - 2 f^{2} + 2 f + 12 \qquad \longleftarrow ^{- 2 f^{2} \text{ terms}}_\text{cancel} \\ 8 f&= 12 \\ f &= \frac{12}{8} \\ f & = \frac{3}{2} \end{aligned}$

Therefore, the value of the variable is $f = \frac{3}{2}$ .

Submit your answer as:

ID is: 3091 Seed is: 2401

Using the proportionality of similar polygons

Polygons $PQT$ and $PRS$ are similar. Compute the value of $f$ . The diagram is not necessarily drawn to scale.

Answer:

f =

$f =$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Similar shapes have proportional lengths. Write a proportion for the sides of the polygons and then solve the equation for $f$ .

STEP: Identify corresponding sides and write a proportion for these sides
[−1 point ⇒ 2 / 3 points left]

We have a diagram with triangles, and we need to figure out the value of the variable $f$ . The question tells us that the triangles are similar, which means that the sides of the shapes are proportional. Therefore, we can write a proportion for the sides of the polygons and solve it for $f$ .

\begin{matrix} \frac{P Q}{Q R} = \frac{P T}{S T} \frac{P Q}{P R} = \frac{P T}{P S} \end{matrix}

$\begin{array}{c}\frac{PQ}{QR} = \frac{PT}{ST} \qquad \frac{PQ}{PR} = \frac{PT}{PS}\end{array}$

STEP: Solve the equation for

f

$f$
[−2 points ⇒ 0 / 3 points left]

\begin{aligned} \frac{P Q}{Q R} & = \frac{P T}{S T} \\ \frac{3 f}{12} & = \frac{- f + 6}{3} \\ (3 f) (3) & = (- f + 6) (12) \\ 9 f & = - 12 f + 72 \\ 21 f & = 72 \\ f & = \frac{72}{21} \\ f & = \frac{24}{7} \end{aligned}

$\begin{aligned} \frac{PQ}{QR} & = \frac{PT}{ST} \\ \frac{3 f}{12} & = \frac{- f + 6}{3} \\ \left( 3 f \right) \left( 3 \right) & = \left(- f + 6 \right) \left( 12 \right) \\ 9 f & = - 12 f + 72 \\ 21 f&= 72 \\ f &= \frac{72}{21} \\ f & = \frac{24}{7} \end{aligned}$

Therefore, the value of the variable is $f = \frac{24}{7}$ .

Submit your answer as:

ID is: 3091 Seed is: 5555

Using the proportionality of similar polygons

In the diagram below, $ABEF\; ||| \;$ $CDFG$ . Compute the value of $y$ . The diagram is not necessarily drawn to scale.

Answer:

y =

$y =$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Similar shapes have proportional lengths. Write a proportion for the sides of the polygons and then solve the equation for $y$ .

STEP: Identify corresponding sides and write a proportion for these sides
[−1 point ⇒ 2 / 3 points left]

We have a diagram with quadrilaterals, and we need to figure out the value of the variable $y$ . The question tells us that the quadrilaterals are similar, which means that the sides of the shapes are proportional. Therefore, we can write a proportion for the sides of the polygons and solve it for $y$ .

\begin{matrix} \frac{A B}{C G} = \frac{E F}{D F} \frac{A B}{E F} = \frac{C G}{D F} \end{matrix}

$\begin{array}{c}\frac{AB}{CG} = \frac{EF}{DF} \qquad \frac{AB}{EF} = \frac{CG}{DF}\end{array}$

STEP: Solve the equation for

y

$y$
[−2 points ⇒ 0 / 3 points left]

We will use the first proportion from above to find the value of $y$ . If you use a different correct proportion and solve it without any mistakes, you will get the correct answer for $y$ (and full marks).

Remember:

\begin{aligned} D F & = E F + D E \\ = - 2 y + 4 + 16 \\ = - 2 y + 20 \end{aligned}

$\begin{aligned} DF &= EF + DE \\ &= - 2 y + 4 + 16 \\ &= - 2 y + 20 \end{aligned}$

So now we can calculate $y$ :

\begin{aligned} \frac{A B}{C G} & = \frac{E F}{D F} \\ \frac{y}{y + 4} & = \frac{- 2 y + 4}{- 2 y + 20} \\ (y) (- 2 y + 20) & = (- 2 y + 4) (y + 4) \\ - 2 y^{2} + 20 y & = - 2 y^{2} - 4 y + 16 ⟵_{cancel}^{- 2 y^{2} terms} \\ 24 y & = 16 \\ y & = \frac{16}{24} \\ y & = \frac{2}{3} \end{aligned}

$\begin{aligned} \frac{AB}{CG} & = \frac{EF}{DF} \\ \frac{y}{y + 4} & = \frac{- 2 y + 4}{- 2 y + 20} \\ \left( y \right) \left( - 2 y + 20 \right) & = \left(- 2 y + 4 \right) \left( y + 4 \right) \\ - 2 y^{2} + 20 y & = - 2 y^{2} - 4 y + 16 \qquad \longleftarrow ^{- 2 y^{2} \text{ terms}}_\text{cancel} \\ 24 y&= 16 \\ y &= \frac{16}{24} \\ y & = \frac{2}{3} \end{aligned}$

Therefore, the value of the variable is $y = \frac{2}{3}$ .

Submit your answer as:

ID is: 4301 Seed is: 3456

Given similar triangles, calculate a side

In the diagram below, $\Delta LNJ \;\vert\vert\vert\; \Delta KNM$ . Also, $JL = 7$ , $MK = 3$ , $MJ = 6$ , and $NM = b$ .

Determine the value of $b$ , giving a reason for your answer.

INSTRUCTION: Give your answer as a fraction, where necessary.

Answer:

b =

$b =$

units

$\text{units}$

one-of

type(numeric.abserror(0.01))

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Since the triangles are similar, you know that the sides are in proportion. Use this fact to create an equation which connects $b$ to the sides that you already know. You may have $b$ more than once in your equation!

STEP: Use the similarity statement to choose the correct pair of ratios
[−2 points ⇒ 2 / 4 points left]

If two triangles are similar, their sides are in proportion. So, when you divide the sides of one triangle with the matching sides in the other triangle, you always get the same number.

TIP: Use the similarity statement to choose the correct pair of ratios by making pairs of matching sides.

We are trying to find $b$ . We can see that $b$ is equal to $NM$ . We can also see that:

\begin{matrix} N J = b + 6 \end{matrix}

$\begin{array}{c} NJ = b + 6 \end{array}$

The similarity is easier to see if we separate the triangles.

We can see from the similarity statement that $\color{blue}{NJ}$ matches with $\color{blue}{NM}$ : $\Delta L{\color{blue}{N}}{\color{blue}{J}} \;\vert\vert\vert\; \Delta K{\color{blue}{N}}{\color{blue}{M}}$ .

We know the lengths of $\color{red}{LJ}$ and $\color{red}{KM}$ . They are a matching pair of sides: $\Delta {\color{red}{L}}N{\color{red}{J}} \;\vert\vert\vert\; \Delta {\color{red}{K}}N{\color{red}{M}}$ .

So, dividing the sides in the big triangle by those in the small triangle, we get

\begin{matrix} \frac{N J}{N M} = \frac{L J}{K M} (Δ L N J | | | Δ K N M) \end{matrix}

$\begin{array}{c} \dfrac{NJ}{NM} = \dfrac{LJ}{KM} \; \left(\Delta LNJ \;\vert\vert\vert\; \Delta KNM\right) \end{array}$

NOTE: We only know that the fractions are equal because the triangles are similar. So, we write this as the reason.

STEP: Solve the equation using inverse operations
[−2 points ⇒ 0 / 4 points left]

First we need to substitute in the values from the diagram. As long as there are only numbers and $b$ -terms, we will be able to solve the equation.

\begin{aligned} \frac{b + 6}{b} & = \frac{7}{3} \\ ∴ & \frac{b + 6}{b} \times b & = \frac{7}{3} \times b \\ ∴ & b + 6 & = \frac{7 b}{3} \\ ∴ & (b + 6) \times 3 & = \frac{7 b}{3} \times 3 \\ ∴ & 3 b + 18 & = 7 b \\ ∴ & 3 b + 18 - 3 b & = 7 b - 3 b \\ ∴ & 18 & = 4 b \\ ∴ & b & = \frac{9}{2} \end{aligned}

$\begin{alignedat}{3} && \dfrac{b + 6}{b} & = \dfrac{7}{3} \\ \therefore && \dfrac{b + 6}{b} \times b & = \dfrac{7}{3} \times b \\ \therefore && b + 6 &= \dfrac{7b}{3} \\ \therefore && (b + 6) \times 3 & = \dfrac{7b}{3} \times 3 \\ \therefore && 3b + 18 & = 7b \\ \therefore && 3b + 18 - 3b & = 7b - 3b \\ \therefore && 18 & = 4b \\ \therefore && b & = \frac{9}{2} \end{alignedat}$

Submit your answer as: and

ID is: 4301 Seed is: 1050

Given similar triangles, calculate a side

In the diagram below, $\Delta PST \;\vert\vert\vert\; \Delta RQT$ . Also, $SP = 11$ , $QR = 6$ , $TQ = 5$ , and $QS = b$ .

Determine the value of $b$ , giving a reason for your answer.

INSTRUCTION: Give your answer as a fraction, where necessary.

Answer:

b =

$b =$

units

$\text{units}$

one-of

type(numeric.abserror(0.01))

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

STEP: Use the similarity statement to choose the correct pair of ratios
[−2 points ⇒ 2 / 4 points left]

If two triangles are similar, their sides are in proportion. So, when you divide the sides of one triangle with the matching sides in the other triangle, you always get the same number.

TIP: Use the similarity statement to choose the correct pair of ratios by making pairs of matching sides.

We are trying to find $b$ . Even though $b$ is not a side in a triangle, it is part of side $TS$ :

\begin{matrix} T S = b + 5 \end{matrix}

$\begin{array}{c} TS = b + 5 \end{array}$

The similarity is easier to see if we separate the triangles.

We can see from the similarity statement that $\color{blue}{ST}$ matches with $\color{blue}{QT}$ : $\Delta P{\color{blue}{S}}{\color{blue}{T}} \;\vert\vert\vert\; \Delta R{\color{blue}{Q}}{\color{blue}{T}}$ .

We know the lengths of $\color{red}{PS}$ and $\color{red}{RQ}$ . They are a matching pair of sides: $\Delta {\color{red}{P}}{\color{red}{S}}T \;\vert\vert\vert\; \Delta {\color{red}{R}}{\color{red}{Q}}T$ .

So, dividing the sides in the big triangle by those in the small triangle, we get

\begin{matrix} \frac{S T}{Q T} = \frac{P S}{R Q} (Δ P S T | | | Δ R Q T) \end{matrix}

$\begin{array}{c} \dfrac{ST}{QT} = \dfrac{PS}{RQ} \; \left(\Delta PST \;\vert\vert\vert\; \Delta RQT\right) \end{array}$

NOTE: We only know that the fractions are equal because the triangles are similar. So, we write this as the reason.

STEP: Solve the equation using inverse operations
[−2 points ⇒ 0 / 4 points left]

First we need to substitute in the values from the diagram. As long as there are only numbers and $b$ -terms, we will be able to solve the equation.

\begin{aligned} \frac{b + 5}{5} & = \frac{11}{6} \\ ∴ & \frac{b + 5}{5} \times 5 & = \frac{11}{6} \times 5 \\ ∴ & b + 5 & = \frac{55}{6} \\ ∴ & b + 5 - 5 & = \frac{55}{6} - 5 \\ ∴ & b & = \frac{25}{6} \end{aligned}

$\begin{alignedat}{3} && \dfrac{b + 5}{5} & = \dfrac{11}{6} \\ \therefore && \dfrac{b + 5}{5} \times 5 & = \dfrac{11}{6} \times 5 \\ \therefore && b + 5 & = \dfrac{55}{6} \\ \therefore && b + 5 - 5 & = \dfrac{55}{6} - 5 \\ \therefore && b & = \frac{25}{6} \end{alignedat}$

Submit your answer as: and

ID is: 4301 Seed is: 4547

Given similar triangles, calculate a side

In the diagram below, $\Delta CBD \;\vert\vert\vert\; \Delta CEA$ . Also, $BD = 8$ , $EA = 3$ , $EB = 7$ , and $CE = y$ .

Determine the value of $y$ , giving a reason for your answer.

INSTRUCTION: Give your answer as a fraction, where necessary.

Answer:

y =

$y =$

units

$\text{units}$

one-of

type(numeric.abserror(0.01))

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Since the triangles are similar, you know that the sides are in proportion. Use this fact to create an equation which connects $y$ to the sides that you already know. You may have $y$ more than once in your equation!

STEP: Use the similarity statement to choose the correct pair of ratios
[−2 points ⇒ 2 / 4 points left]

If two triangles are similar, their sides are in proportion. So, when you divide the sides of one triangle with the matching sides in the other triangle, you always get the same number.

TIP: Use the similarity statement to choose the correct pair of ratios by making pairs of matching sides.

We are trying to find $y$ . We can see that $y$ is equal to $CE$ . We can also see that:

\begin{matrix} C B = y + 7 \end{matrix}

$\begin{array}{c} CB = y + 7 \end{array}$

The similarity is easier to see if we separate the triangles.

We can see from the similarity statement that $\color{blue}{CB}$ matches with $\color{blue}{CE}$ : $\Delta {\color{blue}{C}}{\color{blue}{B}}D \;\vert\vert\vert\; \Delta {\color{blue}{C}}{\color{blue}{E}}A$ .

We know the lengths of $\color{red}{BD}$ and $\color{red}{EA}$ . They are a matching pair of sides: $\Delta C{\color{red}{B}}{\color{red}{D}} \;\vert\vert\vert\; \Delta C{\color{red}{E}}{\color{red}{A}}$ .

So, dividing the sides in the big triangle by those in the small triangle, we get

\begin{matrix} \frac{C B}{C E} = \frac{B D}{E A} (Δ C B D | | | Δ C E A) \end{matrix}

$\begin{array}{c} \dfrac{CB}{CE} = \dfrac{BD}{EA} \; \left(\Delta CBD \;\vert\vert\vert\; \Delta CEA\right) \end{array}$

NOTE: We only know that the fractions are equal because the triangles are similar. So, we write this as the reason.

STEP: Solve the equation using inverse operations
[−2 points ⇒ 0 / 4 points left]

First we need to substitute in the values from the diagram. As long as there are only numbers and $y$ -terms, we will be able to solve the equation.

\begin{aligned} \frac{y + 7}{y} & = \frac{8}{3} \\ ∴ & \frac{y + 7}{y} \times y & = \frac{8}{3} \times y \\ ∴ & y + 7 & = \frac{8 y}{3} \\ ∴ & (y + 7) \times 3 & = \frac{8 y}{3} \times 3 \\ ∴ & 3 y + 21 & = 8 y \\ ∴ & 3 y + 21 - 3 y & = 8 y - 3 y \\ ∴ & 21 & = 5 y \\ ∴ & y & = \frac{21}{5} \end{aligned}

$\begin{alignedat}{3} && \dfrac{y + 7}{y} & = \dfrac{8}{3} \\ \therefore && \dfrac{y + 7}{y} \times y & = \dfrac{8}{3} \times y \\ \therefore && y + 7 &= \dfrac{8y}{3} \\ \therefore && (y + 7) \times 3 & = \dfrac{8y}{3} \times 3 \\ \therefore && 3y + 21 & = 8y \\ \therefore && 3y + 21 - 3y & = 8y - 3y \\ \therefore && 21 & = 5y \\ \therefore && y & = \frac{21}{5} \end{alignedat}$

Submit your answer as: and

ID is: 4304 Seed is: 1103

Given similar triangles, calculate a side

In the diagram below, $\Delta LJK \;|||\; \Delta LMJ$ . Also, $MJ = 24$ , $LJ = 26$ , and $LM = 10$ .

Olatunji has been asked to determine the length of $JK$ . He knows that in order to calculate $JK$ , he first needs to match up the sides of the two similar triangles.

Help Olatunji to match up the sides in the similar triangles.

Answer:

Side in $\Delta LJK$	Matching side in $\Delta LMJ$
$LJ$
$LK$
$JK$

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the similarity statement to choose the correct pairs of sides.

STEP: Use the similarity statement to match the sides
[−2 points ⇒ 0 / 2 points left]

Since $\Delta {\color{red}{L}}{\color{red}{J}}K \;|||\; \Delta {\color{red}{L}}{\color{red}{M}}J$ , we can see that ${\color{red}{LJ}}$ matches with ${\color{red}{LM}}$ . These are the first two letters in the similarity statement.

Also using $\Delta {\color{blue}{L}}J{\color{blue}{K}} \;|||\; \Delta {\color{blue}{L}}M{\color{blue}{J}}$ , we can see that ${\color{blue}{LK}}$ matches with ${\color{blue}{LJ}}$ . These are the first and last letters in the similarity statement.

Finally, we can use $\Delta L{\color{ForestGreen}{J}}{\color{ForestGreen}{K}} \;|||\; \Delta L{\color{ForestGreen}{M}}{\color{ForestGreen}{J}}$ to work out that ${\color{ForestGreen}{JK}}$ matches with ${\color{ForestGreen}{MJ}}$ . These are the second and third letters in the similarity statement.

Visually we can see the matching sides of the triangles:

Submit your answer as: andand

Determine the value of $JK$ .

The diagram is repeated here for your convenience.

INSTRUCTION: Round your answer to two decimal places, if necessary.

Answer: $JK =$ $\text{units}$
one-of
type(numeric.abserror(0.01))
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Use the similarity statement to choose the correct pair of ratios.

STEP: Use the similaity statement to choose the correct pair of ratios
[−2 points ⇒ 1 / 3 points left]

If two triangles are similar, then their sides are in proportion. This means that when you divide the matching sides by each other, you always get the same number.

TIP: Use the similarity statement to choose the correct pair of ratios.

We are looking for $JK$ . So, start by writing $JK$ in the numerator:

$\begin{aligned} \dfrac{JK}{\square} & = \dfrac{\square}{\square} \end{aligned}$

Since $\Delta L{\color{ForestGreen}{J}}{\color{ForestGreen}{K}} \;|||\; \Delta L{\color{ForestGreen}{M}}{\color{ForestGreen}{J}}$ , we can see that ${\color{ForestGreen}{JK}}$ matches with ${\color{ForestGreen}{MJ}}$ .

So we have

$\begin{aligned} \dfrac{JK}{MJ} & = \dfrac{\square}{\square} \end{aligned}$

We were also given information about $LJ$ and $LM$ . Since $\Delta {\color{red}{L}}{\color{red}{J}}K \;|||\; \Delta {\color{red}{L}}{\color{red}{M}}J$ , we can see that ${\color{red}{LJ}}$ matches with ${\color{red}{LM}}$ .

So,

$\begin{aligned} \dfrac{JK}{MJ} & = \dfrac{LJ}{LM} \;\; (\Delta LJK \;|||\; \Delta LMJ) \end{aligned}$

NOTE: We only know that the fractions are equal because the triangles are similar. So, we write this as the reason.

STEP: Substitute in the lengths of the sides and solve the equation
[−1 point ⇒ 0 / 3 points left]

$\begin{aligned} \therefore {\color{ForestGreen}{\dfrac{JK}{24}}} & = {\color{red}{\dfrac{26}{10}}} \\ \therefore JK & = \text{62.4}\ \text{units} \end{aligned}$

Submit your answer as: and

ID is: 4304 Seed is: 5964

Given similar triangles, calculate a side

In the diagram below, $\Delta LJM \;|||\; \Delta LKJ$ . Also, $LJ = 15$ and $LK = 9$ .

Abdulai has been asked to determine the length of $LM$ . He knows that in order to calculate $LM$ , he first needs to match up the sides of the two similar triangles.

Help Abdulai to match up the sides in the similar triangles.

Answer:

Side in $\Delta LJM$	Matching side in $\Delta LKJ$
$LJ$
$LM$
$JM$

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the similarity statement to choose the correct pairs of sides.

STEP: Use the similarity statement to match the sides
[−2 points ⇒ 0 / 2 points left]

Since $\Delta {\color{red}{L}}{\color{red}{J}}M \;|||\; \Delta {\color{red}{L}}{\color{red}{K}}J$ , we can see that ${\color{red}{LJ}}$ matches with ${\color{red}{LK}}$ . These are the first two letters in the similarity statement.

Also using $\Delta {\color{blue}{L}}J{\color{blue}{M}} \;|||\; \Delta {\color{blue}{L}}K{\color{blue}{J}}$ , we can see that ${\color{blue}{LM}}$ matches with ${\color{blue}{LJ}}$ . These are the first and last letters in the similarity statement.

Finally, we can use $\Delta L{\color{ForestGreen}{J}}{\color{ForestGreen}{M}} \;|||\; \Delta L{\color{ForestGreen}{K}}{\color{ForestGreen}{J}}$ to work out that ${\color{ForestGreen}{JM}}$ matches with ${\color{ForestGreen}{KJ}}$ . These are the second and third letters in the similarity statement.

Visually we can see the matching sides of the triangles:

Submit your answer as: andand

Determine the value of $LM$ .

The diagram is repeated here for your convenience.

INSTRUCTION: Round your answer to two decimal places, if necessary.

Answer: $LM =$ $\text{units}$
one-of
type(numeric.abserror(0.01))
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Use the similarity statement to choose the correct pair of ratios.

STEP: Use the similaity statement to choose the correct pair of ratios
[−2 points ⇒ 1 / 3 points left]

If two triangles are similar, then their sides are in proportion. This means that when you divide the matching sides by each other, you always get the same number.

TIP: Use the similarity statement to choose the correct pair of ratios.

We are looking for $LM$ . So, start by writing $LM$ in the numerator:

$\begin{aligned} \dfrac{LM}{\square} & = \dfrac{\square}{\square} \end{aligned}$

Since $\Delta {\color{blue}{L}}J{\color{blue}{M}} \;|||\; \Delta {\color{blue}{L}}K{\color{blue}{J}}$ , we can see that ${\color{blue}{LM}}$ matches with ${\color{blue}{LJ}}$ .

So we have

$\begin{aligned} \dfrac{LM}{LJ} & = \dfrac{\square}{\square} \end{aligned}$

We were also given information about $LJ$ and $LK$ . Since $\Delta {\color{red}{L}}{\color{red}{J}}M \;|||\; \Delta {\color{red}{L}}{\color{red}{K}}J$ , we can see that ${\color{red}{LJ}}$ matches with ${\color{red}{LK}}$ .

So,

$\begin{aligned} \dfrac{LM}{LJ} & = \dfrac{LJ}{LK} \;\; (\Delta LJM \;|||\; \Delta LKJ) \end{aligned}$

NOTE: We only know that the fractions are equal because the triangles are similar. So, we write this as the reason.

STEP: Substitute in the lengths of the sides and solve the equation
[−1 point ⇒ 0 / 3 points left]

$\begin{aligned} \therefore {\color{blue}{\dfrac{LM}{15}}} & = {\color{red}{\dfrac{15}{9}}} \\ \therefore LM & = \text{25}\ \text{units} \end{aligned}$

Submit your answer as: and

ID is: 4304 Seed is: 7470

Given similar triangles, calculate a side

In the diagram below, $\Delta RSP \;|||\; \Delta RQS$ . Also, $QS = 12$ , $RS = 15$ , and $RQ = 9$ .

Anita has been asked to determine the length of $SP$ . She knows that in order to calculate $SP$ , she first needs to match up the sides of the two similar triangles.

Help Anita to match up the sides in the similar triangles.

Answer:

Side in $\Delta RSP$	Matching side in $\Delta RQS$
$RS$
$RP$
$SP$

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the similarity statement to choose the correct pairs of sides.

STEP: Use the similarity statement to match the sides
[−2 points ⇒ 0 / 2 points left]

Since $\Delta {\color{red}{R}}{\color{red}{S}}P \;|||\; \Delta {\color{red}{R}}{\color{red}{Q}}S$ , we can see that ${\color{red}{RS}}$ matches with ${\color{red}{RQ}}$ . These are the first two letters in the similarity statement.

Also using $\Delta {\color{blue}{R}}S{\color{blue}{P}} \;|||\; \Delta {\color{blue}{R}}Q{\color{blue}{S}}$ , we can see that ${\color{blue}{RP}}$ matches with ${\color{blue}{RS}}$ . These are the first and last letters in the similarity statement.

Finally, we can use $\Delta R{\color{ForestGreen}{S}}{\color{ForestGreen}{P}} \;|||\; \Delta R{\color{ForestGreen}{Q}}{\color{ForestGreen}{S}}$ to work out that ${\color{ForestGreen}{SP}}$ matches with ${\color{ForestGreen}{QS}}$ . These are the second and third letters in the similarity statement.

Visually we can see the matching sides of the triangles:

Submit your answer as: andand

Determine the value of $SP$ .

The diagram is repeated here for your convenience.

INSTRUCTION: Round your answer to two decimal places, if necessary.

Answer: $SP =$ $\text{units}$
one-of
type(numeric.abserror(0.01))
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Use the similarity statement to choose the correct pair of ratios.

STEP: Use the similaity statement to choose the correct pair of ratios
[−2 points ⇒ 1 / 3 points left]

If two triangles are similar, then their sides are in proportion. This means that when you divide the matching sides by each other, you always get the same number.

TIP: Use the similarity statement to choose the correct pair of ratios.

We are looking for $SP$ . So, start by writing $SP$ in the numerator:

$\begin{aligned} \dfrac{SP}{\square} & = \dfrac{\square}{\square} \end{aligned}$

Since $\Delta R{\color{ForestGreen}{S}}{\color{ForestGreen}{P}} \;|||\; \Delta R{\color{ForestGreen}{Q}}{\color{ForestGreen}{S}}$ , we can see that ${\color{ForestGreen}{SP}}$ matches with ${\color{ForestGreen}{QS}}$ .

So we have

$\begin{aligned} \dfrac{SP}{QS} & = \dfrac{\square}{\square} \end{aligned}$

We were also given information about $RS$ and $RQ$ . Since $\Delta {\color{red}{R}}{\color{red}{S}}P \;|||\; \Delta {\color{red}{R}}{\color{red}{Q}}S$ , we can see that ${\color{red}{RS}}$ matches with ${\color{red}{RQ}}$ .

So,

$\begin{aligned} \dfrac{SP}{QS} & = \dfrac{RS}{RQ} \;\; (\Delta RSP \;|||\; \Delta RQS) \end{aligned}$

NOTE: We only know that the fractions are equal because the triangles are similar. So, we write this as the reason.

STEP: Substitute in the lengths of the sides and solve the equation
[−1 point ⇒ 0 / 3 points left]

$\begin{aligned} \therefore {\color{ForestGreen}{\dfrac{SP}{12}}} & = {\color{red}{\dfrac{15}{9}}} \\ \therefore SP & = \text{20}\ \text{units} \end{aligned}$

Submit your answer as: and

ID is: 4261 Seed is: 5286

Similarity and congruency notation

Consider the following triangles:

Answer:

Which of the following best describes the relationship between $\Delta PQR$ and $\Delta XYZ$ ?

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The symbol $\; \equiv \;$ is used to indicate that two triangles are congruent. The symbol $\;|||\;$ is used to indicate that two triangles are similar.

STEP: Identify the triangle relationship
[−2 points ⇒ 0 / 2 points left]

Two triangles are similar if they are the same shape, but they can be a different size. The symbol for similarity is $|||$ . There are two ways to know if triangles are similar:

if their sides are in proportion. This means that the sides in one triangle have been multiplied by the same number to get the sides in another triangle.
if all of their matching angles are equal.

Two triangles are congruent if they are exactly the same size and shape. The symbol for congruency is $\equiv$ . There are four ways to know if triangles are congruent: SSS, SAS, SAA, or $90°$ HS.

TIP: When two triangles are equal in all respects, we say that they are congruent and use the symbol for congruency (

\equiv

$\equiv$ ). We never use the equals sign (

=

$=$ ) to compare two triangles.

Two sides and an angle were given in both triangles. But, the angle was not between the two sides, so this is not a case for congruency. This is why we have to prove the included angle in the SAS congruency case. Anyway, the triangles do not look congruent: if we put one on top of the other, they would not match up.

This is also not a case for similarity because the triangles are not the same shape.

Submit your answer as:

None of the above

ID is: 4261 Seed is: 496

Similarity and congruency notation

Consider the following triangles:

Answer:

Which of the following best describes the relationship between $\Delta DEF$ and $\Delta JKL$ ?

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The symbol $\; \equiv \;$ is used to indicate that two triangles are congruent. The symbol $\;|||\;$ is used to indicate that two triangles are similar.

STEP: Identify the triangle relationship
[−2 points ⇒ 0 / 2 points left]

Two triangles are similar if they are the same shape, but they can be a different size. The symbol for similarity is $|||$ . There are two ways to know if triangles are similar:

if their sides are in proportion. This means that the sides in one triangle have been multiplied by the same number to get the sides in another triangle.
if all of their matching angles are equal.

TIP: When two triangles are equal in all respects, we say that they are congruent and use the symbol for congruency (

\equiv

$\equiv$ ). We never use the equals sign (

=

$=$ ) to compare two triangles.

We were given three sides in both triangles. So, the triangles are congruent, and we write:

$\Delta DEF \equiv \Delta JKL$ (SSS)

NOTE: If two triangles are congruent, then they are always also similar. But, being congruent is more special that being similar, so the best way to describe them is to say that they are congruent.

Submit your answer as:

Δ D E F \equiv Δ J K L

$\Delta DEF \equiv \Delta JKL$

ID is: 4261 Seed is: 9067

Similarity and congruency notation

Consider the following triangles:

Answer:

Which of the following best describes the relationship between $\Delta ABC$ and $\Delta DEF$ ?

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The symbol $\; \equiv \;$ is used to indicate that two triangles are congruent. The symbol $\;|||\;$ is used to indicate that two triangles are similar.

STEP: Identify the triangle relationship
[−2 points ⇒ 0 / 2 points left]

Two triangles are similar if they are the same shape, but they can be a different size. The symbol for similarity is $|||$ . There are two ways to know if triangles are similar:

if their sides are in proportion. This means that the sides in one triangle have been multiplied by the same number to get the sides in another triangle.
if all of their matching angles are equal.

TIP: When two triangles are equal in all respects, we say that they are congruent and use the symbol for congruency (

\equiv

$\equiv$ ). We never use the equals sign (

=

$=$ ) to compare two triangles.

The sides in $\Delta DEF$ are each $\text{2}\ \text{units}$ longer than the sides in $\Delta ABC$ . But, this does not mean the sides are in proportion. For the sides to be in proportion, we must multiply by the same number each time (not add). When we divide the pairs of sides in order of length, we do not get the same number each time:

\begin{aligned} \frac{A B}{D E} & = \frac{7}{9} & = 0.77777... \\ \frac{B C}{E F} & = \frac{14.5}{16.5} & = 0.87878... \\ \frac{C A}{F D} & = \frac{12.5}{14.5} & = 0.86206... \end{aligned}

$\begin{alignedat}{2} \dfrac{AB}{DE} & = \dfrac{\text{7}}{\text{9}} && = \text{0.77777...} \\ \dfrac{BC}{EF} & = \dfrac{\text{14.5}}{\text{16.5}} && = \text{0.87878...} \\ \dfrac{CA}{FD} & = \dfrac{\text{12.5}}{\text{14.5}} && = \text{0.86206...} \end{alignedat}$

Even though these triangles look like they are approximately the same shape, they are not similar (or congruent).

Submit your answer as:

None of the above

ID is: 4260 Seed is: 5866

Determine the proportionality factor between similar triangles

The diagram below shows two similar triangles:

All of the sides in $\Delta KJL$ been multiplied by the same number, $k$ , to get the sides in $\Delta PMN$ . What is the value of $k$ ?

INSTRUCTION: Give your answer as an exact decimal or as a fraction.

Answer:

k =

$k =$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Divide the longest side in $\Delta PMN$ by the longest side in $\Delta KJL$ .

STEP: Divide the longest side in

Δ P M N

$\Delta PMN$ by the longest side in

Δ K J L

$\Delta KJL$
[−2 points ⇒ 0 / 2 points left]

We are looking for the number that each matching side in $\Delta KJL$ has been multiplied by, to get the matching side in $\Delta PMN$ .

NOTE: We call this number the proportionality constant,

k

$k$ .

The longest side in $\Delta KJL$ is ${\color{blue}{24}}$ . It must match up with the longest side in $\Delta PMN$ , which is ${\color{blue}{\text{54}}}$ . We will use these sides to determine the multiplier. Let the proportionality constant be $k$ :

\begin{aligned} 24 \times k & = 54 \\ ∴ \frac{24 \times k}{24} & = \frac{54}{24} \\ ∴ k & = 2.25 \end{aligned}

$\begin{aligned} {\color{blue}{24}} \times k & = {\color{blue}{\text{54}}} \\[0.2cm] \therefore \dfrac{24 \times k}{24} & = \dfrac{\text{54}}{24} \\[0.2cm] \therefore k & = \text{2.25} \end{aligned}$

NOTE: In the last step, we divided the longest side in

Δ P M N

$\Delta PMN$ by the longest side in

Δ K J L

$\Delta KJL$ , to find

k

$k$ .

We can check that this value works for the other sides:

The shortest sides, ${\color{red}{KJ}}$ and ${\color{red}{PM}}$ : $\;\;{\color{red}{12}} \times \text{2.25} = {\color{red}{\text{27}}}$
The middle sides, ${\color{ForestGreen}{JL}}$ and ${\color{ForestGreen}{MN}}$ : $\;\;{\color{ForestGreen}{16}} \times \text{2.25} = {\color{ForestGreen}{\text{36}}}$

So, all the sides in $\Delta KJL$ have been multiplied by $\text{2.25}$ to get the sides in $\Delta PMN$ . We call this number the proportionality constant.

Submit your answer as:

ID is: 4260 Seed is: 3237

Determine the proportionality factor between similar triangles

The diagram below shows two similar triangles:

All of the sides in $\Delta JLK$ been multiplied by the same number, $k$ , to get the sides in $\Delta PMN$ . What is the value of $k$ ?

INSTRUCTION: Give your answer as an exact decimal or as a fraction.

Answer:

k =

$k =$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Divide the longest side in $\Delta PMN$ by the longest side in $\Delta JLK$ .

STEP: Divide the longest side in

Δ P M N

$\Delta PMN$ by the longest side in

Δ J L K

$\Delta JLK$
[−2 points ⇒ 0 / 2 points left]

We are looking for the number that each matching side in $\Delta JLK$ has been multiplied by, to get the matching side in $\Delta PMN$ .

NOTE: We call this number the proportionality constant,

k

$k$ .

The longest side in $\Delta JLK$ is ${\color{blue}{14}}$ . It must match up with the longest side in $\Delta PMN$ , which is ${\color{blue}{\text{21}}}$ . We will use these sides to determine the multiplier. Let the proportionality constant be $k$ :

\begin{aligned} 14 \times k & = 21 \\ ∴ \frac{14 \times k}{14} & = \frac{21}{14} \\ ∴ k & = 1.5 \end{aligned}

$\begin{aligned} {\color{blue}{14}} \times k & = {\color{blue}{\text{21}}} \\[0.2cm] \therefore \dfrac{14 \times k}{14} & = \dfrac{\text{21}}{14} \\[0.2cm] \therefore k & = \text{1.5} \end{aligned}$

NOTE: In the last step, we divided the longest side in

Δ P M N

$\Delta PMN$ by the longest side in

Δ J L K

$\Delta JLK$ , to find

k

$k$ .

We can check that this value works for the other sides:

The shortest sides, ${\color{red}{JL}}$ and ${\color{red}{PM}}$ : $\;\;{\color{red}{10}} \times \text{1.5} = {\color{red}{\text{15}}}$
The middle sides, ${\color{ForestGreen}{LK}}$ and ${\color{ForestGreen}{MN}}$ : $\;\;{\color{ForestGreen}{13}} \times \text{1.5} = {\color{ForestGreen}{\text{19.5}}}$

So, all the sides in $\Delta JLK$ have been multiplied by $\text{1.5}$ to get the sides in $\Delta PMN$ . We call this number the proportionality constant.

Submit your answer as:

ID is: 4260 Seed is: 9390

Determine the proportionality factor between similar triangles

The diagram below shows two similar triangles:

All of the sides in $\Delta KJL$ been multiplied by the same number, $k$ , to get the sides in $\Delta NPM$ . What is the value of $k$ ?

INSTRUCTION: Give your answer as an exact decimal or as a fraction.

Answer:

k =

$k =$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Divide the longest side in $\Delta NPM$ by the longest side in $\Delta KJL$ .

STEP: Divide the longest side in

Δ N P M

$\Delta NPM$ by the longest side in

Δ K J L

$\Delta KJL$
[−2 points ⇒ 0 / 2 points left]

We are looking for the number that each matching side in $\Delta KJL$ has been multiplied by, to get the matching side in $\Delta NPM$ .

NOTE: We call this number the proportionality constant,

k

$k$ .

The longest side in $\Delta KJL$ is ${\color{blue}{9}}$ . It must match up with the longest side in $\Delta NPM$ , which is ${\color{blue}{\text{22.5}}}$ . We will use these sides to determine the multiplier. Let the proportionality constant be $k$ :

\begin{aligned} 9 \times k & = 22.5 \\ ∴ \frac{9 \times k}{9} & = \frac{22.5}{9} \\ ∴ k & = 2.5 \end{aligned}

$\begin{aligned} {\color{blue}{9}} \times k & = {\color{blue}{\text{22.5}}} \\[0.2cm] \therefore \dfrac{9 \times k}{9} & = \dfrac{\text{22.5}}{9} \\[0.2cm] \therefore k & = \text{2.5} \end{aligned}$

NOTE: In the last step, we divided the longest side in

Δ N P M

$\Delta NPM$ by the longest side in

Δ K J L

$\Delta KJL$ , to find

k

$k$ .

We can check that this value works for the other sides:

The shortest sides, ${\color{red}{KJ}}$ and ${\color{red}{NP}}$ : $\;\;{\color{red}{5}} \times \text{2.5} = {\color{red}{\text{12.5}}}$
The middle sides, ${\color{ForestGreen}{JL}}$ and ${\color{ForestGreen}{PM}}$ : $\;\;{\color{ForestGreen}{8}} \times \text{2.5} = {\color{ForestGreen}{\text{20}}}$

So, all the sides in $\Delta KJL$ have been multiplied by $\text{2.5}$ to get the sides in $\Delta NPM$ . We call this number the proportionality constant.

Submit your answer as:

ID is: 4294 Seed is: 1891

Similar, congruent, or neither?

Consider the following polygons:

Are the polygons similar, congruent, or neither? Choose the most correct description.

Answer:

The two polygons are:

similar
congruent
not similar or congruent

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Two polygons are similar if they are the same shape, but they may be different sizes. All of their matching angles are equal and all of their matching sides are in proportion.
Two polygons are congruent if they are exactly the same size and shape. All their matching sides and angles are equal. If we cut out two congruent polygons, one could be placed exactly on top of the other.

STEP: Identify if the polygons are similar, congruent, or neither
[−1 point ⇒ 0 / 1 points left]

Two polygons are similar if they are the same shape, but they may be different sizes. All of their matching angles are equal in size and all of their matching sides are in proportion.

Two polygons are congruent if they are exactly the same size and shape. All their matching sides and angles are equal.

The two polygons do not have to be in the same orientation. This means that they can face different directions.

These two quadrilaterals look approximately the same. But, their angles are not exactly equal. So, they cannot be either similar or congruent.

NOTE: In normal English similar means the same in some ways. But the mathematical meaning of the word similar is very specific!

Submit your answer as:

not similar or congruent

ID is: 4294 Seed is: 1206

Similar, congruent, or neither?

Consider the following polygons:

Are the polygons similar, congruent, or neither? Choose the most correct description.

Answer:

The two polygons are:

similar
congruent
not similar or congruent

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Two polygons are similar if they are the same shape, but they may be different sizes. All of their matching angles are equal and all of their matching sides are in proportion.
Two polygons are congruent if they are exactly the same size and shape. All their matching sides and angles are equal. If we cut out two congruent polygons, one could be placed exactly on top of the other.

STEP: Identify if the polygons are similar, congruent, or neither
[−1 point ⇒ 0 / 1 points left]

Two polygons are similar if they are the same shape, but they may be different sizes. All of their matching angles are equal in size and all of their matching sides are in proportion.

Two polygons are congruent if they are exactly the same size and shape. All their matching sides and angles are equal.

The two polygons do not have to be in the same orientation. This means that they can face different directions.

The matching sides and angles in the two polygons are exactly equal.

So, the two polygons are congruent.

NOTE: Congruent polygons are always also similar to each other. But if two polygons are exactly equal, it is more correct to call them congruent.

Submit your answer as:

congruent

ID is: 4294 Seed is: 7911

Similar, congruent, or neither?

Consider the following polygons:

Are the polygons similar, congruent, or neither? Choose the most correct description.

Answer:

The two polygons are:

similar
congruent
not similar or congruent

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Two polygons are similar if they are the same shape, but they may be different sizes. All of their matching angles are equal and all of their matching sides are in proportion.
Two polygons are congruent if they are exactly the same size and shape. All their matching sides and angles are equal. If we cut out two congruent polygons, one could be placed exactly on top of the other.

STEP: Identify if the polygons are similar, congruent, or neither
[−1 point ⇒ 0 / 1 points left]

Two polygons are similar if they are the same shape, but they may be different sizes. All of their matching angles are equal in size and all of their matching sides are in proportion.

Two polygons are congruent if they are exactly the same size and shape. All their matching sides and angles are equal.

The two polygons do not have to be in the same orientation. This means that they can face different directions.

The matching sides and angles in the two polygons are exactly equal.

So, the two polygons are congruent.

NOTE: Congruent polygons are always also similar to each other. But if two polygons are exactly equal, it is more correct to call them congruent.

Submit your answer as:

congruent

ID is: 4256 Seed is: 3792

Identify similar triangles

$\Delta PRQ$ is drawn below.

Which of the following triangles is definitely similar to $\Delta PRQ$ ? Give a reason for the similarity.

Answer:

$\Delta PRQ \;|||\; \Delta$

string

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Triangles are similar if either their:

matching sides are in proportion (sides of $\Delta$ in prop), or
matching angles are equal (equiangular $\Delta$ s).

STEP: Identify the similar triangles
[−1 point ⇒ 1 / 2 points left]

Triangles are similar if either their:

sides are in proportion (sides of $\Delta$ in prop), or
angles are equal (equiangular $\Delta$ s).

We were given three pairs of equal angles in the two triangles, so they are similar.

STEP: Match up the triangle vertices
[−1 point ⇒ 0 / 2 points left]

We must name the triangles in the order that their angles match up.

We can see that the angle at ${\color{red}{P}}$ is equal to the angle at ${\color{red}{E}}$ . So, Point $P$ matches with Point $E$ .

In the same way:

Point ${\color{blue}{R}}$ matches with ${\color{blue}{D}}$ , and
Point ${\color{ForestGreen}{Q}}$ matches with ${\color{ForestGreen}{F}}$ .

So,

$\Delta PRQ \;|||\; \Delta EDF$ (equiangular $\Delta$ s)

NOTE: Equiangular comes from the equal angles in both triangles. It looks similar to equilateral, which means equal lines. But, the two words do not mean the same thing.

Submit your answer as: and

ID is: 4256 Seed is: 1062

Identify similar triangles

$\Delta CAB$ is drawn below.

Which of the following triangles is definitely similar to $\Delta CAB$ ? Give a reason for the similarity.

Answer:

$\Delta CAB \;|||\; \Delta$

string

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Triangles are similar if either their:

matching sides are in proportion (sides of $\Delta$ in prop), or
matching angles are equal (equiangular $\Delta$ s).

STEP: Identify the similar triangles
[−1 point ⇒ 1 / 2 points left]

Triangles are similar if either their:

sides are in proportion (sides of $\Delta$ in prop), or
angles are equal (equiangular $\Delta$ s).

We were given three pairs of equal angles in the two triangles, so they are similar.

STEP: Match up the triangle vertices
[−1 point ⇒ 0 / 2 points left]

We must name the triangles in the order that their angles match up.

We can see that the angle at ${\color{red}{C}}$ is equal to the angle at ${\color{red}{D}}$ . So, Point $C$ matches with Point $D$ .

In the same way:

Point ${\color{blue}{A}}$ matches with ${\color{blue}{E}}$ , and
Point ${\color{ForestGreen}{B}}$ matches with ${\color{ForestGreen}{F}}$ .

So,

$\Delta CAB \;|||\; \Delta DEF$ (equiangular $\Delta$ s)

NOTE: Equiangular comes from the equal angles in both triangles. It looks similar to equilateral, which means equal lines. But, the two words do not mean the same thing.

Submit your answer as: and

ID is: 4256 Seed is: 7399

Identify similar triangles

$\Delta QRP$ is drawn below.

Which of the following triangles is definitely similar to $\Delta QRP$ ? Give a reason for the similarity.

Answer:

$\Delta QRP \;|||\; \Delta$

string

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Triangles are similar if either their:

matching sides are in proportion (sides of $\Delta$ in prop), or
matching angles are equal (equiangular $\Delta$ s).

STEP: Identify the similar triangles
[−1 point ⇒ 1 / 2 points left]

Triangles are similar if either their:

sides are in proportion (sides of $\Delta$ in prop), or
angles are equal (equiangular $\Delta$ s).

We were given three pairs of equal angles in the two triangles, so they are similar.

STEP: Match up the triangle vertices
[−1 point ⇒ 0 / 2 points left]

We must name the triangles in the order that their angles match up.

We can see that the angle at ${\color{red}{Q}}$ is equal to the angle at ${\color{red}{F}}$ . So, Point $Q$ matches with Point $F$ .

In the same way:

Point ${\color{blue}{R}}$ matches with ${\color{blue}{E}}$ , and
Point ${\color{ForestGreen}{P}}$ matches with ${\color{ForestGreen}{D}}$ .

So,

$\Delta QRP \;|||\; \Delta FED$ (equiangular $\Delta$ s)

NOTE: Equiangular comes from the equal angles in both triangles. It looks similar to equilateral, which means equal lines. But, the two words do not mean the same thing.

Submit your answer as: and

ID is: 4285 Seed is: 262

Prove simple similarity with sides

Consider the following triangles:

Prove that $\Delta BAC \;|||\; \Delta EDF$ .

INSTRUCTION: Give your answer as a simplified fraction or decimal.

Answer:

In $\Delta BAC$ and $\Delta EDF$ :

$ED \div$ $=$
$DF \div$ $=$
$EF \div$ $=$

$\therefore \Delta BAC \;|||\; \Delta EDF$ (sides of $\Delta$ in prop)

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

There are two ways of proving that triangles are similar:

Prove that they are equiangular.
Prove that their sides are in proportion.

Which of these strategies is more appropriate for this question?

STEP: Match up sides and find the proportionality constant
[−3 points ⇒ 1 / 4 points left]

There are two ways of proving that triangles are similar:

Prove that they are equiangular.
Prove that their sides are in proportion.

In this question, we have more information about the sides of the triangles than about the angles. So, we will prove that all of the sides are in proportion.

We must show that each side must be multiplied by the same number to make its matching side in the other triangle.

Firstly, we must match up the sides:

The longest sides ( ${\color{red}{ED}}$ and ${\color{red}{BA}}$ ) match up.
The shortest sides ( ${\color{blue}{EF}}$ and ${\color{blue}{BC}}$ ) match up.
${\color{ForestGreen}{FD}}$ and ${\color{ForestGreen}{AC}}$ must match up because they are the only sides left.

Now we will work out what each side in $\Delta BAC$ should be multiplied by:

\begin{aligned} \frac{E D}{B A} & = \frac{20}{16} = 1.25 \\ \frac{D F}{A C} & = \frac{15}{12} = 1.25 \\ \frac{E F}{B C} & = \frac{10}{8} = 1.25 \end{aligned}

$\begin{aligned} \dfrac{ED}{BA} & = \dfrac{\text{20}}{\text{16}} = \text{1.25} \\[0.2cm] \dfrac{DF}{AC} & = \dfrac{\text{15}}{\text{12}} = \text{1.25} \\[0.2cm] \dfrac{EF}{BC} & = \dfrac{\text{10}}{\text{8}} = \text{1.25} \end{aligned}$

This means that any side in $\Delta BAC$ must be multiplied by $\text{1.25}$ to make its matching side in $\Delta EDF$ .

For example, $\text{16} \times \text{1.25} = \text{20}$ .

We call this number the proportionality constant.

If any one of these division sums had given us a different answer, the sides of the triangles would not have been in proportion. Then, the triangles would not be similar.

TIP: If your calculations tell you that the triangles are not similar in a question where you are asked to prove similarity, you probably made a mistake somewhere. Check your working out!

STEP: Prove similarity
[−1 point ⇒ 0 / 4 points left]

Now that we have matched up the sides, and worked out the proportionality constant, we just need to put the information into a formal similarity proof:

In $\Delta BAC$ and $\Delta EDF$ :

$ED \div BA$ $=$ $\dfrac{\text{20}}{\text{16}}$ $=$ $\text{1.25}$
$DF \div AC$ $=$ $\dfrac{\text{15}}{\text{12}}$ $=$ $\text{1.25}$
$EF \div BC$ $=$ $\dfrac{\text{10}}{\text{8}}$ $=$ $\text{1.25}$

$\therefore \Delta BAC \;|||\; \Delta EDF$ (sides of $\Delta$ in prop)

Submit your answer as: andandandandand

ID is: 4285 Seed is: 2870

Prove simple similarity with sides

Consider the following triangles:

Prove that $\Delta QPR \;|||\; \Delta XYZ$ .

INSTRUCTION: Give your answer as a simplified fraction or decimal.

Answer:

In $\Delta QPR$ and $\Delta XYZ$ :

$XY \div$ $=$
$YZ \div$ $=$
$XZ \div$ $=$

$\therefore \Delta QPR \;|||\; \Delta XYZ$ (sides of $\Delta$ in prop)

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

There are two ways of proving that triangles are similar:

Prove that they are equiangular.
Prove that their sides are in proportion.

Which of these strategies is more appropriate for this question?

STEP: Match up sides and find the proportionality constant
[−3 points ⇒ 1 / 4 points left]

There are two ways of proving that triangles are similar:

Prove that they are equiangular.
Prove that their sides are in proportion.

In this question, we have more information about the sides of the triangles than about the angles. So, we will prove that all of the sides are in proportion.

We must show that each side must be multiplied by the same number to make its matching side in the other triangle.

Firstly, we must match up the sides:

The longest sides ( ${\color{red}{XY}}$ and ${\color{red}{QP}}$ ) match up.
The shortest sides ( ${\color{blue}{XZ}}$ and ${\color{blue}{QR}}$ ) match up.
${\color{ForestGreen}{ZY}}$ and ${\color{ForestGreen}{PR}}$ must match up because they are the only sides left.

Now we will work out what each side in $\Delta QPR$ should be multiplied by:

\begin{aligned} \frac{X Y}{Q P} & = \frac{18}{12} = 1.5 \\ \frac{Y Z}{P R} & = \frac{15}{10} = 1.5 \\ \frac{X Z}{Q R} & = \frac{13.5}{9} = 1.5 \end{aligned}

$\begin{aligned} \dfrac{XY}{QP} & = \dfrac{\text{18}}{\text{12}} = \text{1.5} \\[0.2cm] \dfrac{YZ}{PR} & = \dfrac{\text{15}}{\text{10}} = \text{1.5} \\[0.2cm] \dfrac{XZ}{QR} & = \dfrac{\text{13.5}}{\text{9}} = \text{1.5} \end{aligned}$

This means that any side in $\Delta QPR$ must be multiplied by $\text{1.5}$ to make its matching side in $\Delta XYZ$ .

For example, $\text{12} \times \text{1.5} = \text{18}$ .

We call this number the proportionality constant.

If any one of these division sums had given us a different answer, the sides of the triangles would not have been in proportion. Then, the triangles would not be similar.

TIP: If your calculations tell you that the triangles are not similar in a question where you are asked to prove similarity, you probably made a mistake somewhere. Check your working out!

STEP: Prove similarity
[−1 point ⇒ 0 / 4 points left]

Now that we have matched up the sides, and worked out the proportionality constant, we just need to put the information into a formal similarity proof:

In $\Delta QPR$ and $\Delta XYZ$ :

$XY \div QP$ $=$ $\dfrac{\text{18}}{\text{12}}$ $=$ $\text{1.5}$
$YZ \div PR$ $=$ $\dfrac{\text{15}}{\text{10}}$ $=$ $\text{1.5}$
$XZ \div QR$ $=$ $\dfrac{\text{13.5}}{\text{9}}$ $=$ $\text{1.5}$

$\therefore \Delta QPR \;|||\; \Delta XYZ$ (sides of $\Delta$ in prop)

Submit your answer as: andandandandand

ID is: 4285 Seed is: 9881

Prove simple similarity with sides

Consider the following triangles:

Prove that $\Delta ABC \;|||\; \Delta EDF$ .

INSTRUCTION: Give your answer as a simplified fraction or decimal.

Answer:

In $\Delta ABC$ and $\Delta EDF$ :

$ED \div$ $=$
$DF \div$ $=$
$EF \div$ $=$

$\therefore \Delta ABC \;|||\; \Delta EDF$ (sides of $\Delta$ in prop)

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

There are two ways of proving that triangles are similar:

Prove that they are equiangular.
Prove that their sides are in proportion.

Which of these strategies is more appropriate for this question?

STEP: Match up sides and find the proportionality constant
[−3 points ⇒ 1 / 4 points left]

There are two ways of proving that triangles are similar:

Prove that they are equiangular.
Prove that their sides are in proportion.

In this question, we have more information about the sides of the triangles than about the angles. So, we will prove that all of the sides are in proportion.

We must show that each side must be multiplied by the same number to make its matching side in the other triangle.

Firstly, we must match up the sides:

The longest sides ( ${\color{red}{ED}}$ and ${\color{red}{AB}}$ ) match up.
The shortest sides ( ${\color{blue}{EF}}$ and ${\color{blue}{AC}}$ ) match up.
${\color{ForestGreen}{FD}}$ and ${\color{ForestGreen}{BC}}$ must match up because they are the only sides left.

Now we will work out what each side in $\Delta ABC$ should be multiplied by:

\begin{aligned} \frac{E D}{A B} & = \frac{45}{20} = 2.25 \\ \frac{D F}{B C} & = \frac{36}{16} = 2.25 \\ \frac{E F}{A C} & = \frac{27}{12} = 2.25 \end{aligned}

$\begin{aligned} \dfrac{ED}{AB} & = \dfrac{\text{45}}{\text{20}} = \text{2.25} \\[0.2cm] \dfrac{DF}{BC} & = \dfrac{\text{36}}{\text{16}} = \text{2.25} \\[0.2cm] \dfrac{EF}{AC} & = \dfrac{\text{27}}{\text{12}} = \text{2.25} \end{aligned}$

This means that any side in $\Delta ABC$ must be multiplied by $\text{2.25}$ to make its matching side in $\Delta EDF$ .

For example, $\text{20} \times \text{2.25} = \text{45}$ .

We call this number the proportionality constant.

If any one of these division sums had given us a different answer, the sides of the triangles would not have been in proportion. Then, the triangles would not be similar.

TIP: If your calculations tell you that the triangles are not similar in a question where you are asked to prove similarity, you probably made a mistake somewhere. Check your working out!

STEP: Prove similarity
[−1 point ⇒ 0 / 4 points left]

Now that we have matched up the sides, and worked out the proportionality constant, we just need to put the information into a formal similarity proof:

In $\Delta ABC$ and $\Delta EDF$ :

$ED \div AB$ $=$ $\dfrac{\text{45}}{\text{20}}$ $=$ $\text{2.25}$
$DF \div BC$ $=$ $\dfrac{\text{36}}{\text{16}}$ $=$ $\text{2.25}$
$EF \div AC$ $=$ $\dfrac{\text{27}}{\text{12}}$ $=$ $\text{2.25}$

$\therefore \Delta ABC \;|||\; \Delta EDF$ (sides of $\Delta$ in prop)

Submit your answer as: andandandandand

ID is: 4255 Seed is: 2839

Basic deductions from similar triangles

In the diagram below, $\Delta DEF \;|||\; \Delta NML$ , because their sides are in proportion.

What can you work out about the angles in the two triangles?
Answer:
- $\angle D$ $\; = \angle$ $\;\;(\Delta DEF \;|||\; \Delta NML)$
- $\angle F$ $\; = \angle$ $\;\;(\Delta DEF \;|||\; \Delta NML)$
- $\angle E$ $\; = \angle$ $\;\;(\Delta DEF \;|||\; \Delta NML)$
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

If the sides of two triangles are in proportion, then their matching angles are equal. Can you match up the angles of the two triangles?

STEP: Match up the angles
[−3 points ⇒ 0 / 3 points left]

We have been told that $\Delta DEF$ is similar to $\Delta NML$ , because the sides are in proportion. We can see that the all of the sides in $\Delta DEF$ have been multiplied by $\text{3}$ to get the sides in $\Delta NML$ .

If the sides of two triangles are in proportion, then all of their matching angles will be equal. We can identify the matching angles using the similarity statement: $\Delta DEF \;|||\; \Delta NML$ .

So,
- ${\color{blue}{\hat{D}}} = {\color{blue}{\hat{N}}} \;\;(\Delta {\color{blue}{D}}EF \;|||\; \Delta {\color{blue}{N}}ML)$
- ${\color{red}{\hat{E}}} = {\color{red}{\hat{M}}} \;\;(\Delta D{\color{red}{E}}F \;|||\; \Delta N{\color{red}{M}}L)$
- ${\color{ForestGreen}{\hat{F}}} = {\color{ForestGreen}{\hat{L}}} \;\;(\Delta DE{\color{ForestGreen}{F}} \;|||\; \Delta NM{\color{ForestGreen}{L}})$
NOTE: We only know that the angles are equal because the triangles are similar. So, we write this as the reason next to the deduction.

Submit your answer as: andand
In the diagram below, $\Delta PQR \;|||\; \Delta BAC$ , because their angles are equal.

What can you deduce about the sides of the two triangles? Choose the correct option.
Answer:
$\dfrac{PR}{BC} = \dfrac{AC}{QR} = \dfrac{PQ}{BA}$ $\;\;(\Delta PQR \;|||\; \Delta BAC)$

$\dfrac{PR}{BC} = \dfrac{QR}{AC} = \dfrac{PQ}{BA}$ $\;\;(\Delta PQR \;|||\; \Delta BAC)$

$\dfrac{\hat{P}}{\hat{B}} = \dfrac{\hat{Q}}{\hat{A}} = \dfrac{\hat{R}}{\hat{C}}$ $\;\;(\Delta PQR \;|||\; \Delta BAC)$

$\dfrac{PR}{AC} = \dfrac{QR}{BC} = \dfrac{PQ}{BA}$ $\;\;(\Delta PQR \;|||\; \Delta BAC)$
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

If the angles in two triangles are equal, then their sides will be in proportion. Which of the options shows the correct ratios?

STEP: Match up the sides
[−3 points ⇒ 0 / 3 points left]

We have been told that $\Delta PQR$ is similar to $\Delta BAC$ , because the matching angles are equal.

If the matching angles in two triangles are equal, then their sides are in proportion. So, when we divide each side of one triangle by its partner in the other triangle, we always get the same number.

TIP: Use the similarity statement to choose the correct pair of ratios.

Since $\Delta {\color{blue}{P}}{\color{blue}{Q}}R \;|||\; \Delta {\color{blue}{B}}{\color{blue}{A}}C$ , we can see that ${\color{blue}{PQ}}$ matches with ${\color{blue}{BA}}$ .

In the same way, we can see that ${\color{ForestGreen}{PR}}$ matches with ${\color{ForestGreen}{BC}}$ $(\Delta {\color{ForestGreen}{P}}Q{\color{ForestGreen}{R}} \;|||\; \Delta {\color{ForestGreen}{B}}A{\color{ForestGreen}{C}})$ .

Finally, the last sides must also match up. So ${\color{red}{QR}}$ matches with ${\color{red}{AC}}$ $(\Delta P{\color{red}{Q}}{\color{red}{R}} \;|||\; \Delta B{\color{red}{A}}{\color{red}{C}})$ .

TIP: We must always divide the triangle sides in the same order. So, if we put $\Delta PQR$ 's sides in the numerator for the first fraction, then $\Delta PQR$ 's sides must always be in the numerator.

We can also see the matching sides in the diagram:

So, the correct answer is

$\begin{array}{c} \dfrac{PR}{BC} = \dfrac{QR}{AC} = \dfrac{PQ}{BA} \;\; (\Delta PQR \;|||\; \Delta BAC) \end{array}$

NOTE: We only know that the sides are in proportion because the triangles are similar. So, we write this as the reason.

There are lots of correct ways to write the ratios. These are two more correct examples:
1. $\dfrac{BC}{PR} = \dfrac{AC}{QR} = \dfrac{BA}{PQ} \;\; (\Delta PQR \;|||\; \Delta BAC)$
2. $\dfrac{QR}{AC} = \dfrac{PQ}{BA} = \dfrac{PR}{BC} \;\; (\Delta PQR \;|||\; \Delta BAC)$
Submit your answer as:
$\dfrac{PR}{BC} = \dfrac{QR}{AC} = \dfrac{PQ}{BA}$ $\;\;(\Delta PQR \;|||\; \Delta BAC)$

ID is: 4255 Seed is: 4986

Basic deductions from similar triangles

In the diagram below, $\Delta RQP \;|||\; \Delta DEF$ , because their sides are in proportion.

What can you work out about the angles in the two triangles?
Answer:
- $\angle R$ $\; = \angle$ $\;\;(\Delta RQP \;|||\; \Delta DEF)$
- $\angle Q$ $\; = \angle$ $\;\;(\Delta RQP \;|||\; \Delta DEF)$
- $\angle P$ $\; = \angle$ $\;\;(\Delta RQP \;|||\; \Delta DEF)$
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

If the sides of two triangles are in proportion, then their matching angles are equal. Can you match up the angles of the two triangles?

STEP: Match up the angles
[−3 points ⇒ 0 / 3 points left]

We have been told that $\Delta RQP$ is similar to $\Delta DEF$ , because the sides are in proportion. We can see that the all of the sides in $\Delta RQP$ have been multiplied by $\text{2}$ to get the sides in $\Delta DEF$ .

If the sides of two triangles are in proportion, then all of their matching angles will be equal. We can identify the matching angles using the similarity statement: $\Delta RQP \;|||\; \Delta DEF$ .

So,
- ${\color{blue}{\hat{R}}} = {\color{blue}{\hat{D}}} \;\;(\Delta {\color{blue}{R}}QP \;|||\; \Delta {\color{blue}{D}}EF)$
- ${\color{red}{\hat{Q}}} = {\color{red}{\hat{E}}} \;\;(\Delta R{\color{red}{Q}}P \;|||\; \Delta D{\color{red}{E}}F)$
- ${\color{ForestGreen}{\hat{P}}} = {\color{ForestGreen}{\hat{F}}} \;\;(\Delta RQ{\color{ForestGreen}{P}} \;|||\; \Delta DE{\color{ForestGreen}{F}})$
NOTE: We only know that the angles are equal because the triangles are similar. So, we write this as the reason next to the deduction.

Submit your answer as: andand
In the diagram below, $\Delta LMN \;|||\; \Delta BCA$ , because their angles are equal.

What can you deduce about the sides of the two triangles? Choose the correct option.
Answer:
$\dfrac{\hat{B}}{\hat{L}} = \dfrac{\hat{C}}{\hat{M}} = \dfrac{\hat{A}}{\hat{N}}$ $\;\;(\Delta LMN \;|||\; \Delta BCA)$

$\dfrac{MN}{CA} = \dfrac{BC}{LM} = \dfrac{BA}{LN}$ $\;\;(\Delta LMN \;|||\; \Delta BCA)$

$\dfrac{CA}{MN} = \dfrac{BC}{LM} = \dfrac{BA}{LN}$ $\;\;(\Delta LMN \;|||\; \Delta BCA)$

$\dfrac{BA}{MN} = \dfrac{BC}{LM} = \dfrac{CA}{LN}$ $\;\;(\Delta LMN \;|||\; \Delta BCA)$
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

If the angles in two triangles are equal, then their sides will be in proportion. Which of the options shows the correct ratios?

STEP: Match up the sides
[−3 points ⇒ 0 / 3 points left]

We have been told that $\Delta LMN$ is similar to $\Delta BCA$ , because the matching angles are equal.

If the matching angles in two triangles are equal, then their sides are in proportion. So, when we divide each side of one triangle by its partner in the other triangle, we always get the same number.

TIP: Use the similarity statement to choose the correct pair of ratios.

Since $\Delta {\color{blue}{L}}{\color{blue}{M}}N \;|||\; \Delta {\color{blue}{B}}{\color{blue}{C}}A$ , we can see that ${\color{blue}{LM}}$ matches with ${\color{blue}{BC}}$ .

In the same way, we can see that ${\color{ForestGreen}{LN}}$ matches with ${\color{ForestGreen}{BA}}$ $(\Delta {\color{ForestGreen}{L}}M{\color{ForestGreen}{N}} \;|||\; \Delta {\color{ForestGreen}{B}}C{\color{ForestGreen}{A}})$ .

Finally, the last sides must also match up. So ${\color{red}{MN}}$ matches with ${\color{red}{CA}}$ $(\Delta L{\color{red}{M}}{\color{red}{N}} \;|||\; \Delta B{\color{red}{C}}{\color{red}{A}})$ .

TIP: We must always divide the triangle sides in the same order. So, if we put $\Delta BCA$ 's sides in the numerator for the first fraction, then $\Delta BCA$ 's sides must always be in the numerator.

We can also see the matching sides in the diagram:

So, the correct answer is

$\begin{array}{c} \dfrac{CA}{MN} = \dfrac{BC}{LM} = \dfrac{BA}{LN} \;\; (\Delta LMN \;|||\; \Delta BCA) \end{array}$

NOTE: We only know that the sides are in proportion because the triangles are similar. So, we write this as the reason.

There are lots of correct ways to write the ratios. These are two more correct examples:
1. $\dfrac{MN}{CA} = \dfrac{LM}{BC} = \dfrac{LN}{BA} \;\; (\Delta LMN \;|||\; \Delta BCA)$
2. $\dfrac{BC}{LM} = \dfrac{BA}{LN} = \dfrac{CA}{MN} \;\; (\Delta LMN \;|||\; \Delta BCA)$
Submit your answer as:
$\dfrac{CA}{MN} = \dfrac{BC}{LM} = \dfrac{BA}{LN}$ $\;\;(\Delta LMN \;|||\; \Delta BCA)$

ID is: 4255 Seed is: 2115

Basic deductions from similar triangles

In the diagram below, $\Delta FDE \;|||\; \Delta QPR$ , because their sides are in proportion.

What can you work out about the angles in the two triangles?
Answer:
- $\angle F$ $\; = \angle$ $\;\;(\Delta FDE \;|||\; \Delta QPR)$
- $\angle D$ $\; = \angle$ $\;\;(\Delta FDE \;|||\; \Delta QPR)$
- $\angle E$ $\; = \angle$ $\;\;(\Delta FDE \;|||\; \Delta QPR)$
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

If the sides of two triangles are in proportion, then their matching angles are equal. Can you match up the angles of the two triangles?

STEP: Match up the angles
[−3 points ⇒ 0 / 3 points left]

We have been told that $\Delta FDE$ is similar to $\Delta QPR$ , because the sides are in proportion. We can see that the all of the sides in $\Delta FDE$ have been multiplied by $\text{2}$ to get the sides in $\Delta QPR$ .

If the sides of two triangles are in proportion, then all of their matching angles will be equal. We can identify the matching angles using the similarity statement: $\Delta FDE \;|||\; \Delta QPR$ .

So,
- ${\color{blue}{\hat{F}}} = {\color{blue}{\hat{Q}}} \;\;(\Delta {\color{blue}{F}}DE \;|||\; \Delta {\color{blue}{Q}}PR)$
- ${\color{red}{\hat{D}}} = {\color{red}{\hat{P}}} \;\;(\Delta F{\color{red}{D}}E \;|||\; \Delta Q{\color{red}{P}}R)$
- ${\color{ForestGreen}{\hat{E}}} = {\color{ForestGreen}{\hat{R}}} \;\;(\Delta FD{\color{ForestGreen}{E}} \;|||\; \Delta QP{\color{ForestGreen}{R}})$
NOTE: We only know that the angles are equal because the triangles are similar. So, we write this as the reason next to the deduction.

Submit your answer as: andand
In the diagram below, $\Delta ABC \;|||\; \Delta MLN$ , because their angles are equal.

What can you deduce about the sides of the two triangles? Choose the correct option.
Answer:
$\dfrac{\hat{M}}{\hat{A}} = \dfrac{\hat{L}}{\hat{B}} = \dfrac{\hat{N}}{\hat{C}}$ $\;\;(\Delta ABC \;|||\; \Delta MLN)$

$\dfrac{LN}{BC} = \dfrac{AC}{MN} = \dfrac{ML}{AB}$ $\;\;(\Delta ABC \;|||\; \Delta MLN)$

$\dfrac{ML}{BC} = \dfrac{MN}{AC} = \dfrac{LN}{AB}$ $\;\;(\Delta ABC \;|||\; \Delta MLN)$

$\dfrac{LN}{BC} = \dfrac{MN}{AC} = \dfrac{ML}{AB}$ $\;\;(\Delta ABC \;|||\; \Delta MLN)$
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

If the angles in two triangles are equal, then their sides will be in proportion. Which of the options shows the correct ratios?

STEP: Match up the sides
[−3 points ⇒ 0 / 3 points left]

We have been told that $\Delta ABC$ is similar to $\Delta MLN$ , because the matching angles are equal.

If the matching angles in two triangles are equal, then their sides are in proportion. So, when we divide each side of one triangle by its partner in the other triangle, we always get the same number.

TIP: Use the similarity statement to choose the correct pair of ratios.

Since $\Delta {\color{blue}{A}}{\color{blue}{B}}C \;|||\; \Delta {\color{blue}{M}}{\color{blue}{L}}N$ , we can see that ${\color{blue}{AB}}$ matches with ${\color{blue}{ML}}$ .

In the same way, we can see that ${\color{ForestGreen}{AC}}$ matches with ${\color{ForestGreen}{MN}}$ $(\Delta {\color{ForestGreen}{A}}B{\color{ForestGreen}{C}} \;|||\; \Delta {\color{ForestGreen}{M}}L{\color{ForestGreen}{N}})$ .

Finally, the last sides must also match up. So ${\color{red}{BC}}$ matches with ${\color{red}{LN}}$ $(\Delta A{\color{red}{B}}{\color{red}{C}} \;|||\; \Delta M{\color{red}{L}}{\color{red}{N}})$ .

TIP: We must always divide the triangle sides in the same order. So, if we put $\Delta MLN$ 's sides in the numerator for the first fraction, then $\Delta MLN$ 's sides must always be in the numerator.

We can also see the matching sides in the diagram:

So, the correct answer is

$\begin{array}{c} \dfrac{LN}{BC} = \dfrac{MN}{AC} = \dfrac{ML}{AB} \;\; (\Delta ABC \;|||\; \Delta MLN) \end{array}$

NOTE: We only know that the sides are in proportion because the triangles are similar. So, we write this as the reason.

There are lots of correct ways to write the ratios. These are two more correct examples:
1. $\dfrac{BC}{LN} = \dfrac{AC}{MN} = \dfrac{AB}{ML} \;\; (\Delta ABC \;|||\; \Delta MLN)$
2. $\dfrac{MN}{AC} = \dfrac{ML}{AB} = \dfrac{LN}{BC} \;\; (\Delta ABC \;|||\; \Delta MLN)$
Submit your answer as:
$\dfrac{LN}{BC} = \dfrac{MN}{AC} = \dfrac{ML}{AB}$ $\;\;(\Delta ABC \;|||\; \Delta MLN)$

ID is: 4303 Seed is: 1347

Identify similar triangles

In the diagram below, $\hat{K} = 40°$ , and $\hat{J} = \hat{M} = 37°$ .

Identify one triangle that is similar to $\Delta KJN$ .

Answer:

Δ K J N | | | Δ

$\Delta KJN \;|||\; \Delta$

one-of

type(string.nocase)

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Which other triangle in the diagram has the same angles as $\Delta KJN$ ?

STEP: Identify the similar triangle
[−2 points ⇒ 0 / 2 points left]

We can see that $\hat{K}$ is common to $\Delta KML$ and $\Delta KJN$ . We were also told that $\hat{J} = \hat{M}$ .

So, we have two pairs of matching angles in $\Delta KML$ and in $\Delta KJN$ .

We can use sum of angles in both triangles to calculate that $K\hat{L}M = K\hat{N}J = 103°$ .

So all three pairs of matching angles are equal, and $\Delta {\color{red}{K}}{\color{ForestGreen}{M}}{\color{blue}{L}} \;|||\; \Delta {\color{red}{K}}{\color{ForestGreen}{J}}{\color{blue}{N}}$ .

TIP: Whenever we know that two pairs of angles are the same in two triangles, the third pair will automatically also be the same. This is because of the sum of the angles in the two triangles.

Submit your answer as:

ID is: 4303 Seed is: 2589

Identify similar triangles

In the diagram below, $PQ \bot QR$ and $QS \bot PR$ . Also, $\hat{P} = 50°$ and $\hat{R} = 40°$ .

Identify one triangle that is similar to $\Delta QSR$ .

Answer:

Δ Q S R | | | Δ

$\Delta QSR \;|||\; \Delta$

one-of

type(string.nocase)

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Which other triangle in the diagram has the same angles as $\Delta QSR$ ?

STEP: Identify the similar triangle
[−2 points ⇒ 0 / 2 points left]

There are three triangles in the diagram: $\Delta PQR$ , $\Delta QSR$ , and $\Delta PSQ$ . They are all similar to each other.

Looking at $\Delta PQR$ and $\Delta QSR$ , we can see that:

$\hat{R}$ is a common angle.
$P\hat{Q}R = R\hat{S}Q = 90°$ , which was given.
$R\hat{Q}S = \hat{P} = 50°$ , using sum of angles in $\Delta QSR$ .

So, $\Delta {\color{ForestGreen}{P}}{\color{red}{Q}}{\color{blue}{R}} \;|||\; \Delta {\color{ForestGreen}{Q}}{\color{red}{S}}{\color{blue}{R}}$ , because the angles in both triangles are equal.

We could also prove that $\Delta {\color{ForestGreen}{P}}{\color{red}{Q}}{\color{blue}{R}} \;|||\; \Delta {\color{ForestGreen}{P}}{\color{red}{S}}{\color{blue}{Q}}$ .

Since $\Delta QSR$ and $\Delta PSQ$ are both similar to $\Delta PQR$ , they are also similar to each other.

So, either $\Delta PQR$ or $\Delta PSQ$ is a correct answer.

TIP: Whenever we know that two pairs of angles are the same in two triangles, the third pair will automatically also be the same. This is because of the sum of the angles in the two triangles.

Submit your answer as:

ID is: 4303 Seed is: 8966

Identify similar triangles

In the diagram below, $BD \bot DA$ and $DC \bot BA$ . Also, $\hat{B} = 52°$ and $\hat{A} = 38°$ .

Identify one triangle that is similar to $\Delta DCA$ .

Answer:

Δ D C A | | | Δ

$\Delta DCA \;|||\; \Delta$

one-of

type(string.nocase)

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Which other triangle in the diagram has the same angles as $\Delta DCA$ ?

STEP: Identify the similar triangle
[−2 points ⇒ 0 / 2 points left]

There are three triangles in the diagram: $\Delta BDA$ , $\Delta DCA$ , and $\Delta BCD$ . They are all similar to each other.

Looking at $\Delta BDA$ and $\Delta DCA$ , we can see that:

$\hat{A}$ is a common angle.
$B\hat{D}A = A\hat{C}D = 90°$ , which was given.
$A\hat{D}C = \hat{B} = 52°$ , using sum of angles in $\Delta DCA$ .

So, $\Delta {\color{ForestGreen}{B}}{\color{red}{D}}{\color{blue}{A}} \;|||\; \Delta {\color{ForestGreen}{D}}{\color{red}{C}}{\color{blue}{A}}$ , because the angles in both triangles are equal.

We could also prove that $\Delta {\color{ForestGreen}{B}}{\color{red}{D}}{\color{blue}{A}} \;|||\; \Delta {\color{ForestGreen}{B}}{\color{red}{C}}{\color{blue}{D}}$ .

Since $\Delta DCA$ and $\Delta BCD$ are both similar to $\Delta BDA$ , they are also similar to each other.

So, either $\Delta BDA$ or $\Delta BCD$ is a correct answer.

TIP: Whenever we know that two pairs of angles are the same in two triangles, the third pair will automatically also be the same. This is because of the sum of the angles in the two triangles.

Submit your answer as:

ID is: 4363 Seed is: 7583

Definition of similarity and congruency in polygons

Two polygons are congruent if:

Answer:

ATheir angles and sides are equilateral.
BTheir orientation is congruent, and their angles are equilateral.
CThey are equal in size and shape, but they may be in a different orientation.
DAll of their matching angles are equal in size, and all of their matching sides are in proportion.

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Two polygons are similar if they are the same shape, but they may be different sizes. All of their matching angles are equal and all of their matching sides are in proportion.

Two polygons are congruent if they are exactly the same size and shape. All of their matching sides are equal in length and all of their matching angles are equal in size.

STEP: Choose the correct definition
[−2 points ⇒ 0 / 2 points left]

Two polygons are similar if they are the same shape, but they may be different sizes. All of their matching angles are equal, and all of their matching sides are in proportion.

NOTE: In normal English similar means the same in some ways. But the mathematical meaning of the word similar is very specific!

Two polygons are congruent if they are exactly the same size and shape. All of their matching sides and angles are equal.

TIP: Similar or congruent polygons can have a different orientation. This just means that they are facing different directions.

This question asks for the definition of congruency. So, the correct option is C.

NOTE: "Equilateral" means "equal lines". So, angles cannot be equilateral.

Submit your answer as:

CThey are equal in size and shape, but they may be in a different orientation.

ID is: 4363 Seed is: 1539

Definition of similarity and congruency in polygons

Two polygons are similar if:

Answer:

ATheir angles are in proportion.
BTheir angles are equal.
CAll of their matching angles are equal in size, and all of their matching sides are in proportion.
DThey are approximately the same shape.

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Two polygons are similar if they are the same shape, but they may be different sizes. All of their matching angles are equal and all of their matching sides are in proportion.

Two polygons are congruent if they are exactly the same size and shape. All of their matching sides are equal in length and all of their matching angles are equal in size.

STEP: Choose the correct definition
[−2 points ⇒ 0 / 2 points left]

Two polygons are similar if they are the same shape, but they may be different sizes. All of their matching angles are equal, and all of their matching sides are in proportion.

NOTE: In normal English similar means the same in some ways. But the mathematical meaning of the word similar is very specific!

Two polygons are congruent if they are exactly the same size and shape. All of their matching sides and angles are equal.

TIP: Similar or congruent polygons can have a different orientation. This just means that they are facing different directions.

This question asks for the definition of similarity. So, the correct option is C.

NOTE: To be sure that any two polygons are similar, it is not enough to only say that the angles are equal. For example, squares and rectangles always have equal angles, but they are not always similar. It is enough for triangles, but they are very special. That is why you will study similarity in triangles in Grade 9.

Submit your answer as:

CAll of their matching angles are equal in size, and all of their matching sides are in proportion.

ID is: 4363 Seed is: 3877

Definition of similarity and congruency in polygons

Two polygons are congruent if:

Answer:

ATheir angles and sides are equilateral.
BTheir orientation is congruent, and their angles are equilateral.
CThey are exactly the same shape, but they may be different sizes.
DAll of their matching angles are equal in size, and all of their matching sides are equal in length.

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Two polygons are similar if they are the same shape, but they may be different sizes. All of their matching angles are equal and all of their matching sides are in proportion.

Two polygons are congruent if they are exactly the same size and shape. All of their matching sides are equal in length and all of their matching angles are equal in size.

STEP: Choose the correct definition
[−2 points ⇒ 0 / 2 points left]

Two polygons are similar if they are the same shape, but they may be different sizes. All of their matching angles are equal, and all of their matching sides are in proportion.

NOTE: In normal English similar means the same in some ways. But the mathematical meaning of the word similar is very specific!

Two polygons are congruent if they are exactly the same size and shape. All of their matching sides and angles are equal.

TIP: Similar or congruent polygons can have a different orientation. This just means that they are facing different directions.

This question asks for the definition of congruency. So, the correct option is D.

NOTE: "Equilateral" means "equal lines". So, angles cannot be equilateral.

Submit your answer as:

DAll of their matching angles are equal in size, and all of their matching sides are equal in length.

ID is: 4279 Seed is: 6557

Prove similarity with parallel lines

In the diagram below, $NM \parallel KL$ . Also, $NL$ and $MK$ are straight lines.

Prove that $\Delta NMJ \;|||\; \Delta LKJ$ .

Answer:

In $\Delta NMJ$ and $\Delta LKJ$ :

$\angle NJM = \angle LJK$
$\angle M = \angle K$
$\angle N = \angle L$

$\therefore \Delta NMJ \;|||\; \Delta$ .

string

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Two triangles are similar if all of their matching angles are equal.

STEP: Prove that the triangles are equiangular
[−3 points ⇒ 1 / 4 points left]

There are two ways to prove that triangles are similar. We can prove that

their matching sides are in proportion (sides of $\Delta$ in prop), or
their matching angles are equal (equiangular $\Delta$ s).

We have not been given any information about sides, so we will not be able to prove that the sides of these triangles are in proportion. But, we can use our geometry reasons to prove that their matching angles are equal.

TIP: When you are told that two lines are parallel, you should look out for alternate, corresponding, and co-interior angles.

In $\Delta NMJ$ and $\Delta LKJ$ :

${\color{red}{\angle NJM}} = {\color{red}{\angle LJK}}$ (vert opp $\angle$ s equal)
${\color{blue}{\angle M}} = {\color{blue}{\angle K}}$ (alt $\angle$ s; $NM \parallel KL$ )
${\color{ForestGreen}{\angle N}} = {\color{ForestGreen}{\angle L}}$ (alt $\angle$ s; $NM \parallel KL$ )

STEP: Complete the proof by labelling the triangles and giving a reason
[−1 point ⇒ 0 / 4 points left]

We must always match up the triangle vertices on the last line of the congruency proof. This is easier to see when the triangles are drawn separately.

We can see that the angle at $N$ is equal to the angle at $L$ . So, Point ${\color{ForestGreen}{N}}$ matches with Point ${\color{ForestGreen}{L}}$ .

In the same way:

Point ${\color{blue}{M}}$ matches with ${\color{blue}{K}}$ , and
Point ${\color{red}{J}}$ matches with ${\color{red}{J}}$ .

So $\Delta {\color{ForestGreen}{{N}}}{\color{blue}{{M}}}{\color{red}{{J}}} \;|||\; \Delta {\color{ForestGreen}{{L}}}{\color{blue}{{K}}}{\color{red}{{J}}}$ (equiangular $\Delta$ s).

NOTE: You must use the exact reason equiangular $\Delta$ s. This means that each angle in one triangle is equal to an angle in the other triangle.

Submit your answer as: andandandand

ID is: 4279 Seed is: 9149

Prove similarity with parallel lines

In the diagram below, $PT \parallel QR$ .

Prove that $\Delta PTS \;|||\; \Delta QRS$ .

Answer:

In $\Delta PTS$ and $\Delta QRS$ :

$\angle SPT = \angle$ (corresp $\angle$ s; $PT \parallel QR$ )
$\angle STP = \angle$ (corresp $\angle$ s; $PT \parallel QR$ )

$\therefore \Delta PTS \;|||\; \Delta$ .

string

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Two triangles are similar if all of their matching angles are equal.

STEP: Prove that the triangles are equiangular
[−3 points ⇒ 1 / 4 points left]

There are two ways to prove that triangles are similar. We can prove that

their matching sides are in proportion (sides of $\Delta$ in prop), or
their matching angles are equal (equiangular $\Delta$ s).

TIP: When you are told that two lines are parallel, you should look out for alternate, corresponding, and co-interior angles.

In $\Delta PTS$ and $\Delta QRS$ :

${\color{ForestGreen}{\angle SPT}} = {\color{ForestGreen}{\angle SQR}}$ (corresp $\angle$ s; $PT \parallel QR$ )
${\color{blue}{\angle STP}} = {\color{blue}{\angle SRQ}}$ (corresp $\angle$ s; $PT \parallel QR$ )
${\color{red}{\angle S}}$ is common

STEP: Complete the proof by labelling the triangles and giving a reason
[−1 point ⇒ 0 / 4 points left]

We must always match up the triangle vertices on the last line of the congruency proof. This is easier to see when the triangles are drawn separately.

We can see that the angle at $P$ is equal to the angle at $Q$ . So, Point ${\color{ForestGreen}{P}}$ matches with Point ${\color{ForestGreen}{Q}}$ .

In the same way:

Point ${\color{blue}{T}}$ matches with ${\color{blue}{R}}$ , and
Point ${\color{red}{S}}$ matches with ${\color{red}{S}}$ .

So $\Delta {\color{ForestGreen}{{P}}}{\color{blue}{{T}}}{\color{red}{{S}}} \;|||\; \Delta {\color{ForestGreen}{{Q}}}{\color{blue}{{R}}}{\color{red}{{S}}}$ (equiangular $\Delta$ s).

NOTE: You must use the exact reason equiangular $\Delta$ s. This means that each angle in one triangle is equal to an angle in the other triangle.

Submit your answer as: andandandand

ID is: 4279 Seed is: 9682

Prove similarity with parallel lines

In the diagram below, $NL \parallel MJ$ . Also, $NJ$ and $LM$ are straight lines.

Prove that $\Delta NLK \;|||\; \Delta JMK$ .

Answer:

In $\Delta NLK$ and $\Delta JMK$ :

$\angle NKL = \angle$ (vert opp $\angle$ s equal)
$\angle L = \angle$ (alt $\angle$ s; $NL \parallel MJ$ )
$\angle N = \angle$ (alt $\angle$ s; $NL \parallel MJ$ )

$\therefore \Delta NLK \;|||\; \Delta$ .

string

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Two triangles are similar if all of their matching angles are equal.

STEP: Prove that the triangles are equiangular
[−3 points ⇒ 1 / 4 points left]

There are two ways to prove that triangles are similar. We can prove that

their matching sides are in proportion (sides of $\Delta$ in prop), or
their matching angles are equal (equiangular $\Delta$ s).

TIP: When you are told that two lines are parallel, you should look out for alternate, corresponding, and co-interior angles.

In $\Delta NLK$ and $\Delta JMK$ :

${\color{red}{\angle NKL}} = {\color{red}{\angle JKM}}$ (vert opp $\angle$ s equal)
${\color{blue}{\angle L}} = {\color{blue}{\angle M}}$ (alt $\angle$ s; $NL \parallel MJ$ )
${\color{ForestGreen}{\angle N}} = {\color{ForestGreen}{\angle J}}$ (alt $\angle$ s; $NL \parallel MJ$ )

STEP: Complete the proof by labelling the triangles and giving a reason
[−1 point ⇒ 0 / 4 points left]

We must always match up the triangle vertices on the last line of the congruency proof. This is easier to see when the triangles are drawn separately.

We can see that the angle at $N$ is equal to the angle at $J$ . So, Point ${\color{ForestGreen}{N}}$ matches with Point ${\color{ForestGreen}{J}}$ .

In the same way:

Point ${\color{blue}{L}}$ matches with ${\color{blue}{M}}$ , and
Point ${\color{red}{K}}$ matches with ${\color{red}{K}}$ .

So $\Delta {\color{ForestGreen}{{N}}}{\color{blue}{{L}}}{\color{red}{{K}}} \;|||\; \Delta {\color{ForestGreen}{{J}}}{\color{blue}{{M}}}{\color{red}{{K}}}$ (equiangular $\Delta$ s).

NOTE: You must use the exact reason equiangular $\Delta$ s. This means that each angle in one triangle is equal to an angle in the other triangle.

Submit your answer as: andandandand

ID is: 1650 Seed is: 2024

Similar triangles

Given the following diagrams:

Diagram A

Diagram B

Which diagram shows a pair of similar triangles?

Answer: The similar triangles are in Diagram .

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

In maths, the word 'similar' means that two objects have the same shape. Look at the triangles shown and try to determine which pair of triangles are the same shape as each other.

STEP: Compare the triangles and decide which ones are similar
[−2 points ⇒ 0 / 2 points left]

There are two ways we can check for similarity in triangles: using the sides or using the angles.

Angles (AAA): If all three pairs of corresponding angles of two triangles are equal, then the triangles are similar.
Sides (SSS): If all three pairs of corresponding sides of two triangles are in proportion, then the triangles are similar.

We note that we are given information about the angles of all the triangles and so we need to find the diagram that shows a pair of triangles with all three pairs of corresponding angles equal.

We look at the markers for the angles in each triangle and compare these. Diagram A shows a pair of triangles with different angles in each triangle. All six angles are different and there are no pairs of corresponding angles that are equal. Diagram B shows a pair of triangles with all pairs of corresponding angles equal (the same three angle markers are shown in both triangles).

Therefore Diagram B gives a pair of triangles that are similar. When two shapes are similar, we use the symbol $\; ||| \;$ to show it. For example, in Diagram B, $\Delta ACB ||| \Delta DFE$ , which means, 'triangle $ACB$ is similar to triangle $DFE$ .'

The correct choice is B.

Submit your answer as:

ID is: 1650 Seed is: 6206

Similar triangles

Given the following diagrams:

Diagram A

Diagram B

Which diagram shows a pair of similar triangles?

Answer: The similar triangles are in Diagram .

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

In maths, the word 'similar' means that two objects have the same shape. Look at the triangles shown and try to determine which pair of triangles are the same shape as each other.

STEP: Compare the triangles and decide which ones are similar
[−2 points ⇒ 0 / 2 points left]

There are two ways we can check for similarity in triangles: using the sides or using the angles.

Angles (AAA): If all three pairs of corresponding angles of two triangles are equal, then the triangles are similar.
Sides (SSS): If all three pairs of corresponding sides of two triangles are in proportion, then the triangles are similar.

We note that we are given information about the sides of all the triangles and so we need to find a diagram that shows a pair of triangles with all three pairs of corresponding sides in proportion.

We look at the side labels. In Diagram A we note that the three pairs of corresponding sides are in different proportions. In Diagram B we note the three pairs of corresponding sides are in proportion.

The correct choice is B.

Submit your answer as:

ID is: 1650 Seed is: 772

Similar triangles

Given the following diagrams:

Diagram A

Diagram B

Which diagram shows a pair of similar triangles?

Answer: The similar triangles are in Diagram .

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

In maths, the word 'similar' means that two objects have the same shape. Look at the triangles shown and try to determine which pair of triangles are the same shape as each other.

STEP: Compare the triangles and decide which ones are similar
[−2 points ⇒ 0 / 2 points left]

There are two ways we can check for similarity in triangles: using the sides or using the angles.

Angles (AAA): If all three pairs of corresponding angles of two triangles are equal, then the triangles are similar.
Sides (SSS): If all three pairs of corresponding sides of two triangles are in proportion, then the triangles are similar.

We note that we are given information about the angles of all the triangles and so we need to find the diagram that shows a pair of triangles with all three pairs of corresponding angles equal.

We look at the markers for the angles in each triangle and compare these. Diagram A shows a pair of triangles with all pairs of corresponding angles equal (the same three angle markers are shown in both triangles). Diagram B shows a pair of triangles with different angles in each triangle. All six angles are different and there are no pairs of corresponding angles that are equal.

Therefore Diagram A gives a pair of triangles that are similar. When two shapes are similar, we use the symbol $\; ||| \;$ to show it. For example, in Diagram A, $\Delta ACB ||| \Delta DFE$ , which means, 'triangle $ACB$ is similar to triangle $DFE$ .'

The correct choice is A.

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ID is: 4286 Seed is: 7483

Prove simple similarity

Consider the triangles given below:

Select the correct options to prove that $\Delta EFD \;|||\; \Delta NMP$ .

Answer:

In $\Delta EFD$ and $\Delta NMP$ :

	Step	Reason
1.	$\angle F = \angle M = 60°$	(given)
2.	$\angle D = 78°$	(given)
	$= 78°$
	$\therefore$
3.	$\angle N = 42°$	(given)
	$= 42°$
	$\therefore$
	$\therefore \Delta EFD \;\|\|\|\; \Delta NMP$

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

There are two ways of proving similarity:

Prove that the two triangles are equiangular.
Prove that the two triangles have sides in proportion.

Which would be more relevant here?

STEP: Find missing angles
[−4 points ⇒ 1 / 5 points left]

In these two triangles, we have more information about the angles than about the sides. So, we will prove that the two triangles are equiangular.

In order to do this, first we need to find the missing angles.

In $\Delta FDE$ :

$\hat{E} = 180° - 60° - 78°$ $(\text{sum of } \angle \text{s in } \Delta)$
$\therefore \hat{E} = 42°$

In $\Delta MPN$ :

$\hat{P} = 180° - 60° - 42°$ $(\text{sum of } \angle \text{s in } \Delta)$
$\therefore \hat{P} = 78°$

STEP: Complete the proof
[−1 point ⇒ 0 / 5 points left]

Since we have found three pairs of equal angles, we know that the two triangles are similar. Now we need to write our conclusion as a formal proof:

In $\Delta EFD$ and $\Delta NMP$ :

${\color{red}{\hat{F}}} = {\color{red}{\hat{M}}} = 60°$ (given)
$\hat{D} = 78°$ (given)
$\quad$ $\hat{P} = 78°$ (sum of $\angle$ s in $\Delta$ )
$\quad$ $\therefore {\color{ForestGreen}{\hat{D}}} = {\color{ForestGreen}{\hat{P}}}$
$\hat{N} = 42°$ (given)
$\quad$ $\hat{E} = 42°$ (sum of $\angle$ s in $\Delta$ )
$\quad$ $\therefore {\color{blue}{\hat{N}}} = {\color{blue}{\hat{E}}}$

$\therefore \Delta EFD \;|||\; \Delta NMP$ (equiangular $\Delta$ s)

Submit your answer as: andandandandandand

ID is: 4286 Seed is: 4589

Prove simple similarity

Consider the triangles given below:

Select the correct options to prove that $\Delta KLJ \;|||\; \Delta SRQ$ .

Answer:

In $\Delta KLJ$ and $\Delta SRQ$ :

	Step	Reason
1.	$\angle L = \angle R = 48°$	(given)
2.	$\angle J = 53°$	(given)
	$= 53°$
	$\therefore$
3.	$\angle S = 79°$	(given)
	$= 79°$
	$\therefore$
	$\therefore \Delta KLJ \;\|\|\|\; \Delta SRQ$

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

There are two ways of proving similarity:

Prove that the two triangles are equiangular.
Prove that the two triangles have sides in proportion.

Which would be more relevant here?

STEP: Find missing angles
[−4 points ⇒ 1 / 5 points left]

In these two triangles, we have more information about the angles than about the sides. So, we will prove that the two triangles are equiangular.

In order to do this, first we need to find the missing angles.

In $\Delta LJK$ :

$\hat{K} = 180° - 48° - 53°$ $(\text{sum of } \angle \text{s in } \Delta)$
$\therefore \hat{K} = 79°$

In $\Delta RQS$ :

$\hat{Q} = 180° - 48° - 79°$ $(\text{sum of } \angle \text{s in } \Delta)$
$\therefore \hat{Q} = 53°$

STEP: Complete the proof
[−1 point ⇒ 0 / 5 points left]

Since we have found three pairs of equal angles, we know that the two triangles are similar. Now we need to write our conclusion as a formal proof:

In $\Delta KLJ$ and $\Delta SRQ$ :

${\color{red}{\hat{L}}} = {\color{red}{\hat{R}}} = 48°$ (given)
$\hat{J} = 53°$ (given)
$\quad$ $\hat{Q} = 53°$ (sum of $\angle$ s in $\Delta$ )
$\quad$ $\therefore {\color{ForestGreen}{\hat{J}}} = {\color{ForestGreen}{\hat{Q}}}$
$\hat{S} = 79°$ (given)
$\quad$ $\hat{K} = 79°$ (sum of $\angle$ s in $\Delta$ )
$\quad$ $\therefore {\color{blue}{\hat{S}}} = {\color{blue}{\hat{K}}}$

$\therefore \Delta KLJ \;|||\; \Delta SRQ$ (equiangular $\Delta$ s)

Submit your answer as: andandandandandand

ID is: 4286 Seed is: 3114

Prove simple similarity

Consider the triangles given below:

Select the correct options to prove that $\Delta NPM \;|||\; \Delta ZXY$ .

Answer:

In $\Delta NPM$ and $\Delta ZXY$ :

	Step	Reason
1.	$\angle P = \angle X = 86°$	(given)
2.	$\angle M = 40°$	(given)
	$= 40°$
	$\therefore$
3.	$\angle Z = 54°$	(given)
	$= 54°$
	$\therefore$
	$\therefore \Delta NPM \;\|\|\|\; \Delta ZXY$

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

There are two ways of proving similarity:

Prove that the two triangles are equiangular.
Prove that the two triangles have sides in proportion.

Which would be more relevant here?

STEP: Find missing angles
[−4 points ⇒ 1 / 5 points left]

In these two triangles, we have more information about the angles than about the sides. So, we will prove that the two triangles are equiangular.

In order to do this, first we need to find the missing angles.

In $\Delta PMN$ :

$\hat{N} = 180° - 86° - 40°$ $(\text{sum of } \angle \text{s in } \Delta)$
$\therefore \hat{N} = 54°$

In $\Delta XYZ$ :

$\hat{Y} = 180° - 86° - 54°$ $(\text{sum of } \angle \text{s in } \Delta)$
$\therefore \hat{Y} = 40°$

STEP: Complete the proof
[−1 point ⇒ 0 / 5 points left]

Since we have found three pairs of equal angles, we know that the two triangles are similar. Now we need to write our conclusion as a formal proof:

In $\Delta NPM$ and $\Delta ZXY$ :

${\color{red}{\hat{P}}} = {\color{red}{\hat{X}}} = 86°$ (given)
$\hat{M} = 40°$ (given)
$\quad$ $\hat{Y} = 40°$ (sum of $\angle$ s in $\Delta$ )
$\quad$ $\therefore {\color{ForestGreen}{\hat{M}}} = {\color{ForestGreen}{\hat{Y}}}$
$\hat{Z} = 54°$ (given)
$\quad$ $\hat{N} = 54°$ (sum of $\angle$ s in $\Delta$ )
$\quad$ $\therefore {\color{blue}{\hat{Z}}} = {\color{blue}{\hat{N}}}$

$\therefore \Delta NPM \;|||\; \Delta ZXY$ (equiangular $\Delta$ s)

Submit your answer as: andandandandandand

ID is: 4269 Seed is: 3888

Identifying similar and congruent polygons

Polygon $P$ is drawn below:

Which of the following is congruent to Polygon $P$ ?

Answer:

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Two polygons are congruent if they are exactly the same size and shape. If we cut out two congruent polygons, one could be placed exactly on top of the other.

STEP: Choose the congruent polygon
[−2 points ⇒ 0 / 2 points left]

Two polygons are congruent if they are exactly the same size and shape. For two polygons to be congruent, they must have:

the same number of sides,
all of the matching sides equal in length, and
all of the matching angles equal in size.

NOTE: If two polygons are the same shape, but different sizes, then they are similar. But they are not congruent. We could not place one polygon exactly on top of the other!

Option C is exactly the same size and shape as the given polygon. This is true even though they are in different orientations.

TIP: Congruent (or similar) polygons can have a different orientation. This just means that they are facing different directions. If we cut them out and turn them we could still place one directly on top of the other.

Submit your answer as:

ID is: 4269 Seed is: 1658

Identifying similar and congruent polygons

Polygon $P$ is drawn below:

Which of the following is similar to Polygon $P$ ?

Answer:

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Two polygons are similar if they are exactly the same shape, but one might be bigger or smaller than the other.

STEP: Choose the similar polygon
[−2 points ⇒ 0 / 2 points left]

Two polygons are similar if they are exactly the same shape, but one might be bigger or smaller than the other. For two polygons to be similar, they must have:

the same number of sides,
all of the matching sides in proportion, and
all of the matching angles equal in size.

Option C is the same shape as the given polygon, even though it is a different size.

So, Option C is similar to the given polygon.

Submit your answer as:

ID is: 4269 Seed is: 927

Identifying similar and congruent polygons

Polygon $P$ is drawn below:

Which of the following is similar to Polygon $P$ ?

Answer:

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Two polygons are similar if they are exactly the same shape, but one might be bigger or smaller than the other.

STEP: Choose the similar polygon
[−2 points ⇒ 0 / 2 points left]

Two polygons are similar if they are exactly the same shape, but one might be bigger or smaller than the other. For two polygons to be similar, they must have:

the same number of sides,
all of the matching sides in proportion, and
all of the matching angles equal in size.

Option C is the same shape as the given polygon, even though it is a different size.

So, Option C is similar to the given polygon.

Submit your answer as:

ID is: 4313 Seed is: 9709

Calculate a side in similar polygons

The two polygons drawn below are similar.

Determine the value of $y$ .

Answer:

y =

$y=$

units

$\text{units}$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

If two polygons are similar, then their sides are in proportion. This means that every side in the first polygon has been multiplied by the same number to get the matching sides in the second polygon.

STEP: Determine the scale factor
[−2 points ⇒ 0 / 2 points left]

The sides in the first polygon have been multiplied by $\dfrac{\text{1}}{\text{2}}$ to give the matching sides in the second polygon. For example:

${\color{red}{\text{18}}} \times \dfrac{\text{1}}{\text{2}} = {\color{red}{\text{9}}}$
${\color{teal}{\text{14}}} \times \dfrac{\text{1}}{\text{2}} = {\color{teal}{\text{7}}}$

The side labelled $y$ matches with the side equal to ${\color{blue}{\text{8}}}$ in the first polygon. So:

\begin{aligned} 8 \times \frac{1}{2} & = y \\ ∴ y & = 4 units \end{aligned}

$\begin{aligned} {\color{blue}{\text{8}}} \times \dfrac{\text{1}}{\text{2}} &= {\color{blue}{y}} \\ \therefore y &= \text{4}\ \text{units} \end{aligned}$

NOTE: We can see that the sides in the first polygon have all been divided by $\text{2}$ to get the sides in the second polygon. Dividing by

2

$\text{2}$ is the same as multiplying by

\frac{1}{2}

$\dfrac{1}{2}$ .

Submit your answer as:

ID is: 4313 Seed is: 9828

Calculate a side in similar polygons

The two polygons drawn below are similar.

Determine the value of $c$ .

Answer:

c =

$c=$

units

$\text{units}$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

STEP: Determine the scale factor
[−2 points ⇒ 0 / 2 points left]

The sides in the first polygon have been multiplied by $\text{2}$ to give the matching sides in the second polygon. For example:

${\color{red}{\text{10}}} \times \text{2} = {\color{red}{\text{20}}}$
${\color{ForestGreen}{\text{7}}} \times \text{2} = {\color{ForestGreen}{\text{14}}}$

The side labelled $c$ matches with the side equal to ${\color{blue}{\text{8}}}$ in the first polygon. So:

\begin{aligned} 8 \times 2 & = c \\ ∴ c & = 16 units \end{aligned}

$\begin{aligned} {\color{blue}{\text{8}}} \times \text{2} &= {\color{blue}{c}} \\ \therefore c &= \text{16}\ \text{units} \end{aligned}$

Submit your answer as:

ID is: 4313 Seed is: 3182

Calculate a side in similar polygons

The two polygons drawn below are similar.

Determine the value of $x$ .

Answer:

x =

$x=$

units

$\text{units}$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

STEP: Determine the scale factor
[−2 points ⇒ 0 / 2 points left]

The sides in the first polygon have been multiplied by $\text{2}$ to give the matching sides in the second polygon. For example:

${\color{MediumVioletRed}{\text{5}}} \times \text{2} = {\color{MediumVioletRed}{\text{10}}}$
${\color{red}{\text{4}}} \times \text{2} = {\color{red}{\text{8}}}$

The side labelled $x$ matches with the side equal to ${\color{ForestGreen}{\text{9}}}$ in the first polygon. So:

\begin{aligned} 9 \times 2 & = x \\ ∴ x & = 18 units \end{aligned}

$\begin{aligned} {\color{ForestGreen}{\text{9}}} \times \text{2} &= {\color{ForestGreen}{x}} \\ \therefore x &= \text{18}\ \text{units} \end{aligned}$

Submit your answer as:

ID is: 4362 Seed is: 7656

Similar statements: true or false

Consider the following statements and decide whether they are true or false.

Answer:

All equilateral triangles are similar to each other.
All isosceles triangles are similar to each other.

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Two polygons are similar if they are the same shape, but they may be different sizes. All of their matching angles are equal and all of their matching sides are in proportion.

Are all equilateral triangles the same shape? What about isosceles triangles?

STEP: Are all equilateral triangles the same shape?
[−2 points ⇒ 2 / 4 points left]

Two polygons are similar if they are the same shape, but they may be different sizes. All of their matching angles are equal and all of their matching sides are in proportion.

All the angles in equilateral triangles are always $\text{60}\text{°}$ . Their sides are always in proportion: they are always the same shape, but they may be different sizes.

✔ All equilateral triangles are similar to each other.

This statement is true.

STEP: Are all isosceles triangles the same shape?
[−2 points ⇒ 0 / 4 points left]

Isosceles triangles always have two equal angles. But, the sizes of those angles can change. So, not all isosceles triangles are similar to each other, because they are not always the same shape.

✘ All isosceles triangles are similar to each other.

This statement is false.

NOTE: An example that disagrees with the statement is called a counter-example. We only need one counter-example to prove that a statement is false. This is because if it is false for one example, then we cannot say that it is true for all - so it must be false.

Submit your answer as: and

ID is: 4362 Seed is: 5903

Similar statements: true or false

Consider the following statements and decide whether they are true or false.

Answer:

All regular pentagons are similar to each other.
All pentagons are similar to each other.

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Two polygons are similar if they are the same shape, but they may be different sizes. All of their matching angles are equal and all of their matching sides are in proportion.

Are all regular pentagons the same shape? What about pentagons?

STEP: Are all regular pentagons the same shape?
[−2 points ⇒ 2 / 4 points left]

Two polygons are similar if they are the same shape, but they may be different sizes. All of their matching angles are equal and all of their matching sides are in proportion.

All the angles in regular pentagons are always $\text{108}\text{°}$ . Their sides are always in proportion: they are always the same shape, but they may be different sizes.

✔ All regular pentagons are similar to each other.

This statement is true.

STEP: Are all pentagons the same shape?
[−2 points ⇒ 0 / 4 points left]

Pentagons always have five sides. But, the angles in a pentagon can be different, and the sides can be different lengths. So, not all pentagons are similar to each other, because they are not always the same shape.

✘ All pentagons are similar to each other.

This statement is false.

Submit your answer as: and

ID is: 4362 Seed is: 2078

Similar statements: true or false

Consider the following statements and decide whether they are true or false.

Answer:

All equilateral triangles are similar to each other.
All equilateral triangles are congruent to each other.

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Two polygons are similar if they are the same shape, but they may be different sizes. All of their matching angles are equal and all of their matching sides are in proportion. Are all equilateral triangles the same shape?

Two polygons are congruent if they are exactly the same size and shape. All their matching sides and angles are equal. Are all equilateral triangles exactly the same size and shape?

STEP: Are all equilateral triangles the same shape?
[−2 points ⇒ 2 / 4 points left]

Two polygons are similar if they are the same shape, but they may be different sizes. All of their matching angles are equal and all of their matching sides are in proportion.

All the angles in equilateral triangles are always $\text{60}\text{°}$ . Their sides are always in proportion: they are always the same shape, but they may be different sizes.

✔ All equilateral triangles are similar to each other.

This statement is true.

STEP: Are all equilateral triangles the same size and shape?
[−2 points ⇒ 0 / 4 points left]

Two polygons are congruent if they are exactly the same size and shape. All their matching sides and angles are equal. If we cut out two congruent polygons, one could be placed exactly on top of the other.

Equilateral triangles can be different sizes. We can see this in the diagrams above. So, not all equilateral triangles are congruent to each other.

✘ All equilateral triangles are congruent to each other.

This statement is false.

Submit your answer as: and

ID is: 4321 Seed is: 6085

Similarity or congruency?

Consider the diagram below and answer the questions that follow:

Lehan needs to prove that $\Delta MNL \;|||\; \Delta XYZ$ . He has already started, and his incomplete proof is written below:

In $\Delta MNL$ and $\Delta XYZ$ :
1. $\hat{M} = \hat{X}$ (given)
  ...
Answer:
How should Lehan complete his proof? Choose the best option.
Option A
2. $\hat{N} = \hat{Y}$ (given)
3. $\hat{L} = \hat{Z}$ (given)
$\therefore \Delta MNL \;|||\; \Delta XYZ$ (equiangular $\Delta$ s)

Option B
2. $\hat{N} = \hat{Y}$ (given)
3. $\hat{L} = 59°$ (sum of $\angle$ s in $\Delta$ )
Also $\hat{Z} = 59°$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{L} = \hat{Z}$
$\therefore \Delta MNL \;|||\; \Delta XYZ$ (equiangular $\Delta$ s)

Option C
2. $\hat{N} = \hat{Y}$ (given)
3. $\dfrac{MN}{XZ} = \dfrac{10}{10} = 1$
$\therefore \Delta MNL \;|||\; \Delta XYZ$ (sides of $\Delta$ in prop)

Option D
2. $\hat{N} = \hat{Y}$ (given)
3. $MN = ZX$ (given)
$\therefore \Delta MNL \equiv \Delta XYZ$ (SAA)
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Two triangles are similar if they are the same shape, but they may be a different size. Their angles will be equal and their sides will be in proportion.

Two triangles are congruent if they are exactly the same size and shape. Their matching sides and angles will all be equal.

STEP: Choose the correct proof
[−4 points ⇒ 0 / 4 points left]

✘ Option A

This "proof" makes a mistake on Step 3. $\hat{L}$ is equal to $\hat{Z}$ because of sum of $\angle$ s in $\Delta$ . It was not given.

✔ Option B

This option uses sum of $\angle$ s in $\Delta$ correctly to prove that there are three pairs of matching angles in the triangles.

✘ Option C

To prove that two triangles are similar, you must either prove that all of the sides are in proportion, or that all the matching angles are equal . This proof tries to do a mixture of these two, which is not allowed.

✘ Option D

We cannot prove that the triangles are congruent, even though we have been given a pair of equal sides and two pairs of equal angles. This is because the angles are not in the same positions relative to the given side. Even though the triangles look almost the same size, the information we have been given actually proves that they are slightly different sizes.

TIP: The question told us to prove that the triangles are similar (using the symbol $|||$ ). You should always read the question statement very carefully.

The similarity is easier to see when the triangles are in the same orientation:

The full proof is shown below:

In $\color{gray}{\Delta MNL}$ and $\color{gray}{\Delta XYZ}$ :
1. $\color{gray}{\hat{M} = \hat{X}}$ (given)
2. $\hat{N} = \hat{Y}$ (given)
3. $\hat{L} = 59°$ (sum of $\angle$ s in $\Delta$ )
  Also $\hat{Z} = 59°$ (sum of $\angle$ s in $\Delta$ )
  $\therefore \hat{L} = \hat{Z}$
$\therefore \Delta MNL \;|||\; \Delta XYZ$ (equiangular $\Delta$ s)

Submit your answer as:
Option B
2. $\hat{N} = \hat{Y}$ (given)
3. $\hat{L} = 59°$ (sum of $\angle$ s in $\Delta$ )
Also $\hat{Z} = 59°$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{L} = \hat{Z}$
$\therefore \Delta MNL \;|||\; \Delta XYZ$ (equiangular $\Delta$ s)
Hence determine the length of $XY$ , giving a reason for your answer.

INSTRUCTION: Do not round off your answer.

Answer: $XY =$
one-of
type(numeric.abserror(0.01))
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Use the similarity statement to choose the correct pair of ratios.

STEP: Use the similarity statement to choose the correct pair of ratios
[−2 points ⇒ 1 / 3 points left]

If two triangles are similar, their sides are in proportion. This means that when you divide the matching sides by each other, you always get the same number.

TIP: Use the similarity statement to choose the correct pair of ratios.

We are looking for $XY$ . So, start by writing $XY$ in the numerator:

$\begin{aligned} \dfrac{XY}{\Box} = \dfrac{\Box}{\Box} \end{aligned}$

Since $\Delta {\color{teal}{M}}{\color{teal}{N}}L \;|||\; \Delta {\color{teal}{X}}{\color{teal}{Y}}Z$ , we can see that ${\color{teal}{XY}}$ matches with ${\color{teal}{MN}}$ .

So we have:

$\begin{aligned} \dfrac{XY}{MN} = \dfrac{\Box}{\Box} \end{aligned}$

We were also given information about $ZX$ and $ML$ . Since $\Delta {\color{orange}{M}}N{\color{orange}{L}} \;|||\; \Delta {\color{orange}{X}}Y{\color{orange}{Z}}$ , we can see that ${\color{orange}{ZX}}$ matches with ${\color{orange}{ML}}$ .

So,

$\begin{aligned} \dfrac{XY}{MN} = \dfrac{ZX}{ML} \; (\Delta MNL \;|||\; \Delta XYZ) \end{aligned}$

NOTE: We only know that the fractions are equal because the triangles are similar. So we write this as the reason. Remember, sides of $\Delta$ in prop is used to prove that triangles are similar, not to determine side lengths.

STEP: Solve the equation using inverse operations
[−1 point ⇒ 0 / 3 points left]

$\begin{aligned} \dfrac{{\color{teal}{XY}}}{{\color{teal}{10}}} & = \dfrac{{\color{orange}{10}}}{{\color{orange}{8}}} \\ \therefore \dfrac{XY}{10} \times 10 & = \dfrac{10}{8} \times 10 \\ \therefore XY & = \frac{25}{2}\\ & = \text{12.5} \end{aligned}$

Submit your answer as: and

ID is: 4321 Seed is: 6902

Similarity or congruency?

Consider the diagram below and answer the questions that follow:

Jabulile needs to prove that $\Delta STU \;|||\; \Delta EDF$ . She has already started, and her incomplete proof is written below:

In $\Delta STU$ and $\Delta EDF$ :
1. $\hat{S} = \hat{E}$ (given)
  ...
Answer:
How should Jabulile complete her proof? Choose the best option.
Option A
2. $\hat{D} = 84°$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{T} = \hat{D}$
3. $\dfrac{ST}{EF} = \dfrac{12}{12} = 1$
$\therefore \Delta STU \;|||\; \Delta EDF$ (sides of $\Delta$ in prop)

Option B
2. $\hat{U} = 53°$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{U} = \hat{F}$
3. $ST = FE$ (given)
$\therefore \Delta STU \;|||\; \Delta EDF$ (SAA)

Option C
2. $\hat{U} = 53°$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{U} = \hat{F}$
3. Also, $\hat{D} = 84°$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{T} = \hat{D}$
$\therefore \Delta STU \;|||\; \Delta EDF$ (equiangular $\Delta$ s)

Option D
2. $\hat{T} = \hat{D}$ (given)
3. $\hat{U} = 53°$ (sum of $\angle$ s in $\Delta$ )
Also, $\hat{F} = 53°$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{U} = \hat{F}$
$\therefore \Delta STU \;|||\; \Delta EDF$ (equiangular $\Delta$ s)
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Two triangles are similar if they are the same shape, but they may be a different size. Their angles will be equal and their sides will be in proportion.

Two triangles are congruent if they are exactly the same size and shape. Their matching sides and angles will all be equal.

STEP: Choose the correct proof
[−4 points ⇒ 0 / 4 points left]

✘ Option A

To prove that two triangles are similar, you must either prove that all of the sides are in proportion, or that all the matching angles are equal . This proof tries to do a mixture of these two, which is not allowed.

✘ Option B

We cannot prove that the triangles are congruent, even though we can prove a pair of equal sides and two pairs of equal angles. This is because the angles are not in the same positions relative to the given side. The information we have been given actually proves that the triangles are slightly different sizes.

✔ Option C

This option uses sum of $\angle$ s in $\Delta$ correctly to prove that there are three pairs of matching angles in the triangles.

✘ Option D

This "proof" makes mistakes on Step 2 and Step 3. We were not given that $\hat{T} = \hat{D}$ ; we must prove this using sum of $\angle$ s in $\Delta$ . Also, on Step 3, we know that $\hat{F} = 53°$ because this was given , not because of sum of $\angle$ s in $\Delta$ .

TIP: The question told us to prove that the triangles are similar (using the symbol $|||$ ). You should always read the question statement very carefully.

The similarity is easier to see when the triangles are in the same orientation:

The full proof is shown below:

In $\color{gray}{\Delta STU}$ and $\color{gray}{\Delta EDF}$ :
1. $\color{gray}{\hat{S} = \hat{E}}$ (given)
2. $\hat{U} = 53°$ (sum of $\angle$ s in $\Delta$ )
  $\therefore \hat{U} = \hat{F}$
3. Also, $\hat{D} = 84°$ (sum of $\angle$ s in $\Delta$ )
  $\therefore \hat{T} = \hat{D}$
$\therefore \Delta STU \;|||\; \Delta EDF$ (equiangular $\Delta$ s)

Submit your answer as:
Option C
2. $\hat{U} = 53°$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{U} = \hat{F}$
3. Also, $\hat{D} = 84°$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{T} = \hat{D}$
$\therefore \Delta STU \;|||\; \Delta EDF$ (equiangular $\Delta$ s)
Hence determine the length of $ED$ , giving a reason for your answer.

INSTRUCTION: Do not round off your answer.

Answer: $ED =$
one-of
type(numeric.abserror(0.01))
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Use the similarity statement to choose the correct pair of ratios.

STEP: Use the similarity statement to choose the correct pair of ratios
[−2 points ⇒ 1 / 3 points left]

If two triangles are similar, their sides are in proportion. This means that when you divide the matching sides by each other, you always get the same number.

TIP: Use the similarity statement to choose the correct pair of ratios.

We are looking for $ED$ . So, start by writing $ED$ in the numerator:

$\begin{aligned} \dfrac{ED}{\Box} = \dfrac{\Box}{\Box} \end{aligned}$

Since $\Delta {\color{teal}{S}}{\color{teal}{T}}U \;|||\; \Delta {\color{teal}{E}}{\color{teal}{D}}F$ , we can see that ${\color{teal}{ED}}$ matches with ${\color{teal}{ST}}$ .

So we have:

$\begin{aligned} \dfrac{ED}{ST} = \dfrac{\Box}{\Box} \end{aligned}$

We were also given information about $FE$ and $SU$ . Since $\Delta {\color{orange}{S}}T{\color{orange}{U}} \;|||\; \Delta {\color{orange}{E}}D{\color{orange}{F}}$ , we can see that ${\color{orange}{FE}}$ matches with ${\color{orange}{SU}}$ .

So,

$\begin{aligned} \dfrac{ED}{ST} = \dfrac{FE}{SU} \; (\Delta STU \;|||\; \Delta EDF) \end{aligned}$

NOTE: We only know that the fractions are equal because the triangles are similar. So we write this as the reason. Remember, sides of $\Delta$ in prop is used to prove that triangles are similar, not to determine side lengths.

STEP: Solve the equation using inverse operations
[−1 point ⇒ 0 / 3 points left]

$\begin{aligned} \dfrac{{\color{teal}{ED}}}{{\color{teal}{12}}} & = \dfrac{{\color{orange}{12}}}{{\color{orange}{15}}} \\ \therefore \dfrac{ED}{12} \times 12 & = \dfrac{12}{15} \times 12 \\ \therefore ED & = \frac{48}{5}\\ & = \text{9.6} \end{aligned}$

Submit your answer as: and

ID is: 4321 Seed is: 9269

Similarity or congruency?

Consider the diagram below and answer the questions that follow:

Danladi needs to prove that $\Delta ABC \;|||\; \Delta XYZ$ . He has already started, and his incomplete proof is written below:

In $\Delta ABC$ and $\Delta XYZ$ :
1. $\hat{A} = \hat{X}$ (given)
  ...
Answer:
How should Danladi complete his proof? Choose the best option.
Option A
2. $\hat{C} = 82°$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{C} = \hat{Z}$
3. Also, $\hat{Y} = 44°$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{B} = \hat{Y}$
$\therefore \Delta ABC \;|||\; \Delta XYZ$ (equiangular $\Delta$ s)

Option B
2. $\hat{C} = 82°$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{C} = \hat{Z}$
3. $AB = ZX$ (given)
$\therefore \Delta ABC \;|||\; \Delta XYZ$ (SAA)

Option C
2. $\hat{Y} = 44°$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{B} = \hat{Y}$
3. $\dfrac{AB}{XZ} = \dfrac{7}{7} = 1$
$\therefore \Delta ABC \;|||\; \Delta XYZ$ (sides of $\Delta$ in prop)

Option D
2. $\hat{B} = \hat{Y}$ (given)
3. $\hat{C} = 82°$ (sum of $\angle$ s in $\Delta$ )
Also, $\hat{Z} = 82°$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{C} = \hat{Z}$
$\therefore \Delta ABC \;|||\; \Delta XYZ$ (equiangular $\Delta$ s)
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Two triangles are similar if they are the same shape, but they may be a different size. Their angles will be equal and their sides will be in proportion.

Two triangles are congruent if they are exactly the same size and shape. Their matching sides and angles will all be equal.

STEP: Choose the correct proof
[−4 points ⇒ 0 / 4 points left]

✔ Option A

This option uses sum of $\angle$ s in $\Delta$ correctly to prove that there are three pairs of matching angles in the triangles.

✘ Option B

We cannot prove that the triangles are congruent, even though we can prove a pair of equal sides and two pairs of equal angles. This is because the angles are not in the same positions relative to the given side. The information we have been given actually proves that the triangles are slightly different sizes.

✘ Option C

To prove that two triangles are similar, you must either prove that all of the sides are in proportion, or that all the matching angles are equal . This proof tries to do a mixture of these two, which is not allowed.

✘ Option D

This "proof" makes mistakes on Step 2 and Step 3. We were not given that $\hat{B} = \hat{Y}$ ; we must prove this using sum of $\angle$ s in $\Delta$ . Also, on Step 3, we know that $\hat{Z} = 82°$ because this was given , not because of sum of $\angle$ s in $\Delta$ .

TIP: The question told us to prove that the triangles are similar (using the symbol $|||$ ). You should always read the question statement very carefully.

The similarity is easier to see when the triangles are in the same orientation:

The full proof is shown below:

In $\color{gray}{\Delta ABC}$ and $\color{gray}{\Delta XYZ}$ :
1. $\color{gray}{\hat{A} = \hat{X}}$ (given)
2. $\hat{C} = 82°$ (sum of $\angle$ s in $\Delta$ )
  $\therefore \hat{C} = \hat{Z}$
3. Also, $\hat{Y} = 44°$ (sum of $\angle$ s in $\Delta$ )
  $\therefore \hat{B} = \hat{Y}$
$\therefore \Delta ABC \;|||\; \Delta XYZ$ (equiangular $\Delta$ s)

Submit your answer as:
Option A
2. $\hat{C} = 82°$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{C} = \hat{Z}$
3. Also, $\hat{Y} = 44°$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{B} = \hat{Y}$
$\therefore \Delta ABC \;|||\; \Delta XYZ$ (equiangular $\Delta$ s)
Hence determine the length of $XY$ , giving a reason for your answer.

INSTRUCTION: Do not round off your answer.

Answer: $XY =$
one-of
type(numeric.abserror(0.01))
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Use the similarity statement to choose the correct pair of ratios.

STEP: Use the similarity statement to choose the correct pair of ratios
[−2 points ⇒ 1 / 3 points left]

If two triangles are similar, their sides are in proportion. This means that when you divide the matching sides by each other, you always get the same number.

TIP: Use the similarity statement to choose the correct pair of ratios.

We are looking for $XY$ . So, start by writing $XY$ in the numerator:

$\begin{aligned} \dfrac{XY}{\Box} = \dfrac{\Box}{\Box} \end{aligned}$

Since $\Delta {\color{teal}{A}}{\color{teal}{B}}C \;|||\; \Delta {\color{teal}{X}}{\color{teal}{Y}}Z$ , we can see that ${\color{teal}{XY}}$ matches with ${\color{teal}{AB}}$ .

So we have:

$\begin{aligned} \dfrac{XY}{AB} = \dfrac{\Box}{\Box} \end{aligned}$

We were also given information about $ZX$ and $AC$ . Since $\Delta {\color{orange}{A}}B{\color{orange}{C}} \;|||\; \Delta {\color{orange}{X}}Y{\color{orange}{Z}}$ , we can see that ${\color{orange}{ZX}}$ matches with ${\color{orange}{AC}}$ .

So,

$\begin{aligned} \dfrac{XY}{AB} = \dfrac{ZX}{AC} \; (\Delta ABC \;|||\; \Delta XYZ) \end{aligned}$

NOTE: We only know that the fractions are equal because the triangles are similar. So we write this as the reason. Remember, sides of $\Delta$ in prop is used to prove that triangles are similar, not to determine side lengths.

STEP: Solve the equation using inverse operations
[−1 point ⇒ 0 / 3 points left]

$\begin{aligned} \dfrac{{\color{teal}{XY}}}{{\color{teal}{7}}} & = \dfrac{{\color{orange}{7}}}{{\color{orange}{5}}} \\ \therefore \dfrac{XY}{7} \times 7 & = \dfrac{7}{5} \times 7 \\ \therefore XY & = \frac{49}{5}\\ & = \text{9.8} \end{aligned}$

Submit your answer as: and

2. Scale factors

ID is: 4290 Seed is: 1518

Consequences of similarity with calculations

In the diagram below, $\Delta ABC \;|||\; \Delta FED$ .

Determine the size of $\hat{F}$ , in terms of $x$ .
Answer:
$\hat{A} =$ $°$
$\therefore \hat{F} =$ $°$

expression
expression
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

In similar triangles,
1. angles are equal, and
2. sides are in proportion.
You do not have enough information to calculate $\hat{F}$ in one step. Are there any other angles that you can calculate?

Then, you can use the fact that the triangles are similar.

STEP: Calculate the missing angle in $\Delta ABC$
[−2 points ⇒ 2 / 4 points left]

We cannot calculate $\hat{F}$ in one step, because we do not have enough information in $\Delta FED$ .

But, we do have enough information in $\Delta ABC$ to calculate $\hat{A}$ . Knowing the size of $\hat{A}$ will help us to calculate $\hat{F}$ .

In $\Delta ABC$ :

$\begin{aligned} \hat{A} + (x + 10°) + 90° & = 180° \; (\text{sum of } \angle \text{s in } \Delta) \end{aligned}$

$\begin{aligned} \therefore \hat{A} & = 180° - 90° - x - 10° \\ \therefore \hat{A} & = 80° - x \end{aligned}$

STEP: Compare matching angles in $\Delta FED$
[−2 points ⇒ 0 / 4 points left]

Since $\Delta ABC \;|||\; \Delta FED$ , we know that the two triangles are equiangular and have sides in proportion.

We are looking for an angle. So, we will use the fact that matching angles are equal between similar triangles to calculate $\hat{F}$ .

We were told that $\Delta ABC \;|||\; \Delta FED$ . So, ${\color{red}{\hat{A}}}$ and ${\color{red}{\hat{F}}}$ are matching angles, because their letters come first. Similarly, ${\color{blue}{\hat{B}}}$ and ${\color{blue}{\hat{E}}}$ are matching angles. Finally, ${\color{ForestGreen}{\hat{C}}}$ and ${\color{ForestGreen}{\hat{D}}}$ are matching angles.

$\hat{A}$ matches $\hat{F}$ . So,

$\begin{aligned} \hat{F} & = 80° - x \;\; (\Delta ABC \;|||\; \Delta FED) \end{aligned}$

Submit your answer as: andandand
Now, use the same diagram to determine the length of $AC$ .
Answer:
$FD =$ $\text{units}$
$\therefore AC =$ $\text{units}$

numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

In similar triangles,
1. angles are equal, and
2. sides are in proportion.
You do not have enough information to calculate $AC$ in one step. Are there any other sides that you can calculate?

Then, you can use the fact that the triangles are similar.

STEP: Calculate the missing side in $\Delta FED$
[−2 points ⇒ 2 / 4 points left]

We cannot calculate $AC$ in one step, because we do not have enough information in $\Delta ABC$ .

But, we do have enough information in $\Delta FED$ to calculate $FD$ . Then we can use the fact that the triangles are similar to calculate $AC$ .

In $\Delta FED$ :

$\begin{aligned} \text{20}^{2} & = \text{16}^{2} + FD^{2} \; \; (\text{Pythagoras}) \\ \therefore \text{144} & = FD^{2} \\ \therefore \text{12} & = FD \end{aligned}$

NOTE: We can only use the theorem of Pythagoras because $\Delta FED$ is right-angled!

STEP: Compare matching sides in $\Delta ABC$
[−2 points ⇒ 0 / 4 points left]

Since $\Delta ABC \;|||\; \Delta FED$ , we know that the two triangles are equiangular and have sides in proportion.

We are looking for a side. So, we will use the fact that sides in similar triangles are in proportion to calculate $AC$ .

We were told that $\Delta ABC \;|||\; \Delta FED$ . So, ${\color{blue}{AB}}$ and ${\color{blue}{FE}}$ are matching sides, because their letters came first. In the same way, ${\color{red}{BC}}$ and ${\color{red}{ED}}$ are matching sides because their letters came last. And ${\color{ForestGreen}{AC}}$ and ${\color{ForestGreen}{FD}}$ are matching sides, because they are the only sides left.

We know the lengths of $ED$ and $BC$ , so we will use them to calculate the value of the proportionality constant.

We want to know what to multiply the sides of $\Delta FED$ by to get the sides of $\Delta ABC$ . So, we divide the known side of $\Delta ABC$ by the known side of $\Delta FED$ .

$\begin{aligned} k & = \dfrac{BC}{DE} \\ & = \dfrac{8}{16} \\ & = \text{0.5} \end{aligned}$

The sides are in proportion, so when we multiply any side in $\Delta FED$ by $\text{0.5}$ , we will get the matching side in $\Delta ABC$ .

$\begin{aligned} \therefore AC & = FD \times \text{0.5} \;\; (\Delta ABC \;|||\; \Delta FED) \\ \therefore AC & = \text{12} \times \text{0.5} \\ & = \text{6} \end{aligned}$

Submit your answer as: andandand

ID is: 4290 Seed is: 4253

Consequences of similarity with calculations

In the diagram below, $\Delta QRS \;|||\; \Delta XZY$ .

Determine the size of $\hat{X}$ , in terms of $x$ .
Answer:
$\hat{Q} =$ $°$
$\therefore \hat{X} =$ $°$

expression
expression
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

In similar triangles,
1. angles are equal, and
2. sides are in proportion.
You do not have enough information to calculate $\hat{X}$ in one step. Are there any other angles that you can calculate?

Then, you can use the fact that the triangles are similar.

STEP: Calculate the missing angle in $\Delta QRS$
[−2 points ⇒ 2 / 4 points left]

We cannot calculate $\hat{X}$ in one step, because we do not have enough information in $\Delta XZY$ .

But, we do have enough information in $\Delta QRS$ to calculate $\hat{Q}$ . Knowing the size of $\hat{Q}$ will help us to calculate $\hat{X}$ .

In $\Delta QRS$ :

$\begin{aligned} \hat{Q} + (x + 10°) + 90° & = 180° \; (\text{sum of } \angle \text{s in } \Delta) \end{aligned}$

$\begin{aligned} \therefore \hat{Q} & = 180° - 90° - x - 10° \\ \therefore \hat{Q} & = 80° - x \end{aligned}$

STEP: Compare matching angles in $\Delta XZY$
[−2 points ⇒ 0 / 4 points left]

Since $\Delta QRS \;|||\; \Delta XZY$ , we know that the two triangles are equiangular and have sides in proportion.

We are looking for an angle. So, we will use the fact that matching angles are equal between similar triangles to calculate $\hat{X}$ .

We were told that $\Delta QRS \;|||\; \Delta XZY$ . So, ${\color{red}{\hat{Q}}}$ and ${\color{red}{\hat{X}}}$ are matching angles, because their letters come first. Similarly, ${\color{blue}{\hat{R}}}$ and ${\color{blue}{\hat{Z}}}$ are matching angles. Finally, ${\color{ForestGreen}{\hat{S}}}$ and ${\color{ForestGreen}{\hat{Y}}}$ are matching angles.

$\hat{Q}$ matches $\hat{X}$ . So,

$\begin{aligned} \hat{X} & = 80° - x \;\; (\Delta QRS \;|||\; \Delta XZY) \end{aligned}$

Submit your answer as: andandand
Now, use the same diagram to determine the length of $QS$ .
Answer:
$XY =$ $\text{units}$
$\therefore QS =$ $\text{units}$

numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

In similar triangles,
1. angles are equal, and
2. sides are in proportion.
You do not have enough information to calculate $QS$ in one step. Are there any other sides that you can calculate?

Then, you can use the fact that the triangles are similar.

STEP: Calculate the missing side in $\Delta XZY$
[−2 points ⇒ 2 / 4 points left]

We cannot calculate $QS$ in one step, because we do not have enough information in $\Delta QRS$ .

But, we do have enough information in $\Delta XZY$ to calculate $XY$ . Then we can use the fact that the triangles are similar to calculate $QS$ .

In $\Delta XZY$ :

$\begin{aligned} \text{30}^{2} & = \text{24}^{2} + XY^{2} \; \; (\text{Pythagoras}) \\ \therefore \text{324} & = XY^{2} \\ \therefore \text{18} & = XY \end{aligned}$

NOTE: We can only use the theorem of Pythagoras because $\Delta XZY$ is right-angled!

STEP: Compare matching sides in $\Delta QRS$
[−2 points ⇒ 0 / 4 points left]

Since $\Delta QRS \;|||\; \Delta XZY$ , we know that the two triangles are equiangular and have sides in proportion.

We are looking for a side. So, we will use the fact that sides in similar triangles are in proportion to calculate $QS$ .

We were told that $\Delta QRS \;|||\; \Delta XZY$ . So, ${\color{blue}{QR}}$ and ${\color{blue}{XZ}}$ are matching sides, because their letters came first. In the same way, ${\color{red}{RS}}$ and ${\color{red}{ZY}}$ are matching sides because their letters came last. And ${\color{ForestGreen}{QS}}$ and ${\color{ForestGreen}{XY}}$ are matching sides, because they are the only sides left.

We know the lengths of $ZY$ and $RS$ , so we will use them to calculate the value of the proportionality constant.

We want to know what to multiply the sides of $\Delta XZY$ by to get the sides of $\Delta QRS$ . So, we divide the known side of $\Delta QRS$ by the known side of $\Delta XZY$ .

$\begin{aligned} k & = \dfrac{RS}{YZ} \\ & = \dfrac{12}{24} \\ & = \text{0.5} \end{aligned}$

The sides are in proportion, so when we multiply any side in $\Delta XZY$ by $\text{0.5}$ , we will get the matching side in $\Delta QRS$ .

$\begin{aligned} \therefore QS & = XY \times \text{0.5} \;\; (\Delta QRS \;|||\; \Delta XZY) \\ \therefore QS & = \text{18} \times \text{0.5} \\ & = \text{9} \end{aligned}$

Submit your answer as: andandand

ID is: 4290 Seed is: 8318

Consequences of similarity with calculations

In the diagram below, $\Delta ABC \;|||\; \Delta FED$ .

Determine the size of $\hat{F}$ , in terms of $x$ .
Answer:
$\hat{A} =$ $°$
$\therefore \hat{F} =$ $°$

expression
expression
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

In similar triangles,
1. angles are equal, and
2. sides are in proportion.
You do not have enough information to calculate $\hat{F}$ in one step. Are there any other angles that you can calculate?

Then, you can use the fact that the triangles are similar.

STEP: Calculate the missing angle in $\Delta ABC$
[−2 points ⇒ 2 / 4 points left]

We cannot calculate $\hat{F}$ in one step, because we do not have enough information in $\Delta FED$ .

But, we do have enough information in $\Delta ABC$ to calculate $\hat{A}$ . Knowing the size of $\hat{A}$ will help us to calculate $\hat{F}$ .

In $\Delta ABC$ :

$\begin{aligned} \hat{A} + (x + 10°) + 90° & = 180° \; (\text{sum of } \angle \text{s in } \Delta) \end{aligned}$

$\begin{aligned} \therefore \hat{A} & = 180° - 90° - x - 10° \\ \therefore \hat{A} & = 80° - x \end{aligned}$

STEP: Compare matching angles in $\Delta FED$
[−2 points ⇒ 0 / 4 points left]

Since $\Delta ABC \;|||\; \Delta FED$ , we know that the two triangles are equiangular and have sides in proportion.

We are looking for an angle. So, we will use the fact that matching angles are equal between similar triangles to calculate $\hat{F}$ .

We were told that $\Delta ABC \;|||\; \Delta FED$ . So, ${\color{red}{\hat{A}}}$ and ${\color{red}{\hat{F}}}$ are matching angles, because their letters come first. Similarly, ${\color{blue}{\hat{B}}}$ and ${\color{blue}{\hat{E}}}$ are matching angles. Finally, ${\color{ForestGreen}{\hat{C}}}$ and ${\color{ForestGreen}{\hat{D}}}$ are matching angles.

$\hat{A}$ matches $\hat{F}$ . So,

$\begin{aligned} \hat{F} & = 80° - x \;\; (\Delta ABC \;|||\; \Delta FED) \end{aligned}$

Submit your answer as: andandand
Now, use the same diagram to determine the length of $AC$ .
Answer:
$FD =$ $\text{units}$
$\therefore AC =$ $\text{units}$

numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

In similar triangles,
1. angles are equal, and
2. sides are in proportion.
You do not have enough information to calculate $AC$ in one step. Are there any other sides that you can calculate?

Then, you can use the fact that the triangles are similar.

STEP: Calculate the missing side in $\Delta FED$
[−2 points ⇒ 2 / 4 points left]

We cannot calculate $AC$ in one step, because we do not have enough information in $\Delta ABC$ .

But, we do have enough information in $\Delta FED$ to calculate $FD$ . Then we can use the fact that the triangles are similar to calculate $AC$ .

In $\Delta FED$ :

$\begin{aligned} \text{34}^{2} & = \text{30}^{2} + FD^{2} \; \; (\text{Pythagoras}) \\ \therefore \text{256} & = FD^{2} \\ \therefore \text{16} & = FD \end{aligned}$

NOTE: We can only use the theorem of Pythagoras because $\Delta FED$ is right-angled!

STEP: Compare matching sides in $\Delta ABC$
[−2 points ⇒ 0 / 4 points left]

Since $\Delta ABC \;|||\; \Delta FED$ , we know that the two triangles are equiangular and have sides in proportion.

We are looking for a side. So, we will use the fact that sides in similar triangles are in proportion to calculate $AC$ .

We were told that $\Delta ABC \;|||\; \Delta FED$ . So, ${\color{blue}{AB}}$ and ${\color{blue}{FE}}$ are matching sides, because their letters came first. In the same way, ${\color{red}{BC}}$ and ${\color{red}{ED}}$ are matching sides because their letters came last. And ${\color{ForestGreen}{AC}}$ and ${\color{ForestGreen}{FD}}$ are matching sides, because they are the only sides left.

We know the lengths of $ED$ and $BC$ , so we will use them to calculate the value of the proportionality constant.

We want to know what to multiply the sides of $\Delta FED$ by to get the sides of $\Delta ABC$ . So, we divide the known side of $\Delta ABC$ by the known side of $\Delta FED$ .

$\begin{aligned} k & = \dfrac{BC}{DE} \\ & = \dfrac{15}{30} \\ & = \text{0.5} \end{aligned}$

The sides are in proportion, so when we multiply any side in $\Delta FED$ by $\text{0.5}$ , we will get the matching side in $\Delta ABC$ .

$\begin{aligned} \therefore AC & = FD \times \text{0.5} \;\; (\Delta ABC \;|||\; \Delta FED) \\ \therefore AC & = \text{16} \times \text{0.5} \\ & = \text{8} \end{aligned}$

Submit your answer as: andandand

3. Practical applications

ID is: 4314 Seed is: 5346

Applications of similarity

Usman wants to find out the height of a school building. Usman is $\text{1.57}\ \text{m}$ tall. The length of Usman's shadow changes throughout the day.

At $\text{9}\ \text{am}$ , Usman stands in the shadow of the school building. The shadows of the school building and Usman form triangles. Both triangles contain the same angle, as shown in the diagram below.

At $\text{9}\ \text{am}$ , Usman's shadow is $\text{2.72}\ \text{m}$ long and the school building's shadow is $\text{6.53}\ \text{m}$ long. Complete the proof to help Usman work out the height of the school building.
Answer:
In $\Delta UVW$ and $\Delta ZYW$ :
1. is common.
2. $\angle U = \angle WZY =$ $°$ (because the school building and Usman are both perpendicular to the ground)
3. $\angle V = \angle WYZ =$ $°$ (sum of $\angle$ s in $\Delta$ )
$\therefore \Delta UVW \;|||\; \Delta ZYW$

numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

It may help you to redraw the diagram with the two triangles separate from each other. Which pairs of angles are equal between the two triangles, and why?

STEP: Prove similarity
[−4 points ⇒ 0 / 4 points left]

TIP: In real-life questions like this one, we can always assume that vertical objects like school buildings and human beings are perpendicular to the ground. We also assume that the ground is perfectly horizontal!

Once we know that the angles at $V\hat{U}W$ and $Y\hat{Z}W$ are $90°$ , we can use sum of $\angle$ s in $\Delta$ to work out expressions for the angles at $U\hat{V}W$ and $Z\hat{Y}W$ .

In $\Delta UVW$ :

$\begin{aligned} {\color{MediumVioletRed}{\hat{U}}} + {\color{orange}{\hat{V}}} + {\color{teal}{\hat{W}}} & = 180° \; (\text{sum of } \angle \text{s in } \Delta) \\ \therefore 90° + \hat{V} + \theta & = 180° \\ \therefore 90° + \hat{V} + \theta {\color{blue}{-\; 90°}} {\color{red}{- \;\theta}} & = 180° {\color{blue}{-\; 90°}} {\color{red}{-\; \theta}} \\ \therefore \hat{V} & = 90° - \theta \end{aligned}$

Similarly, in $\Delta ZYW$ :

$\begin{aligned} {\color{MediumVioletRed}{\hat{Z}}} + {\color{orange}{\hat{Y}}} + {\color{teal}{\hat{W}}} & = 180° \; (\text{sum of } \angle \text{s in } \Delta) \\ \therefore 90° + \hat{Y} + \theta & = 180° \\ \therefore \hat{Y} & = 90° - \theta \end{aligned}$

So, even though we don't know what $\theta$ is, we can say for certain that $\hat{V} = \hat{Y}$ , because they are both equal to $90° - \theta$ .

For example, if $\theta = 30°$ , $\hat{V}$ and $\hat{Y}$ would both be $60°$ . If $\theta = 50°$ , $\hat{V}$ and $\hat{Y}$ would both be $40°$ . Since they both have the same algebraic expression, they will both have the same value no matter what value we substitute for $\theta$ .

The completed proof is:

In $\Delta UVW$ and $\Delta ZYW$ :
1. $\angle W$ is common.
2. $\angle U = \angle Z = 90°$ (because the school building and Usman are both perpendicular to the ground)
3. $\angle V = \angle Y = 90° - \theta$ (sum of $\angle$ s in $\Delta$ )
$\therefore \Delta UVW \;|||\; \Delta ZYW$ (equiangular $\Delta$ s)

Submit your answer as: andandand
We have been told that at $\text{9}\ \text{am}$ , Usman's shadow is $\text{2.72}\ \text{m}$ long and the school building's shadow is $\text{6.53}\ \text{m}$ long. We have also proved that $\Delta UVW \;|||\; \Delta ZYW$ .

Hence, work out the height of the school building.

INSTRUCTION: Round your answer to two decimal places.

Answer:
Height of the school building $=$ $\text{m}$ $\;$ $(\Delta UVW \;|||\; \Delta ZYW)$

one-of
type(numeric.abserror(0.01))
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Since the two triangles are similar, their sides are in proportion.

STEP: Use sides in proportion to create and solve an equation
[−3 points ⇒ 0 / 3 points left]

Since we have proved that $\Delta UVW \;|||\; \Delta ZYW$ , their sides must be in proportion. We can use the ratio of sides to set up an equation.

Since the height of the school building ( $UV$ ) is what we want to find out, we will put it in the numerator. It matches with Usman's height ( $ZY$ ) in the smaller triangle.

$\begin{aligned} \dfrac{UV}{\text{1.57}} & = \dfrac{\square}{\square} \end{aligned}$

In the same way, the school building's shadow ( $UW$ ) matches with Usman's shadow ( $ZW$ ) in the smaller triangle.

$\begin{aligned} \dfrac{UV}{\text{1.57}} & = \dfrac{\text{6.53}}{\text{2.72}} \end{aligned}$

Hence,

$\begin{aligned} \dfrac{UV}{\text{1.57}} {\color{ForestGreen}{\, \times \, \text{1.57}}} & = \dfrac{\text{6.53}}{\text{2.72}} {\color{ForestGreen}{\, \times \, \text{1.57}}} \\ \therefore UV & = \text{3.76915...}\ \text{m} \\ & \approx \text{3.77}\ \text{m} \end{aligned}$

Submit your answer as:

ID is: 4314 Seed is: 7536

Applications of similarity

Eric wants to find out the height of a lamp post. Eric is $\text{1.53}\ \text{m}$ tall. The length of Eric's shadow changes throughout the day.

At $\text{8}\ \text{am}$ , Eric stands in the shadow of the lamp post. The shadows of the lamp post and Eric form triangles. Both triangles contain the same angle, as shown in the diagram below.

At $\text{8}\ \text{am}$ , Eric's shadow is $\text{2.65}\ \text{m}$ long and the lamp post's shadow is $\text{6.1}\ \text{m}$ long. Complete the proof to help Eric work out the height of the lamp post.
Answer:
In $\Delta ABC$ and $\Delta FEC$ :
1. is common.
2. $\angle A = \angle CFE =$ $°$ (because the lamp post and Eric are both perpendicular to the ground)
3. $\angle B = \angle CEF =$ $°$ (sum of $\angle$ s in $\Delta$ )
$\therefore \Delta ABC \;|||\; \Delta FEC$

numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

It may help you to redraw the diagram with the two triangles separate from each other. Which pairs of angles are equal between the two triangles, and why?

STEP: Prove similarity
[−4 points ⇒ 0 / 4 points left]

TIP: In real-life questions like this one, we can always assume that vertical objects like lamp posts and human beings are perpendicular to the ground. We also assume that the ground is perfectly horizontal!

Once we know that the angles at $B\hat{A}C$ and $E\hat{F}C$ are $90°$ , we can use sum of $\angle$ s in $\Delta$ to work out expressions for the angles at $A\hat{B}C$ and $F\hat{E}C$ .

In $\Delta ABC$ :

$\begin{aligned} {\color{MediumVioletRed}{\hat{A}}} + {\color{orange}{\hat{B}}} + {\color{teal}{\hat{C}}} & = 180° \; (\text{sum of } \angle \text{s in } \Delta) \\ \therefore 90° + \hat{B} + \theta & = 180° \\ \therefore 90° + \hat{B} + \theta {\color{blue}{-\; 90°}} {\color{red}{- \;\theta}} & = 180° {\color{blue}{-\; 90°}} {\color{red}{-\; \theta}} \\ \therefore \hat{B} & = 90° - \theta \end{aligned}$

Similarly, in $\Delta FEC$ :

$\begin{aligned} {\color{MediumVioletRed}{\hat{F}}} + {\color{orange}{\hat{E}}} + {\color{teal}{\hat{C}}} & = 180° \; (\text{sum of } \angle \text{s in } \Delta) \\ \therefore 90° + \hat{E} + \theta & = 180° \\ \therefore \hat{E} & = 90° - \theta \end{aligned}$

So, even though we don't know what $\theta$ is, we can say for certain that $\hat{B} = \hat{E}$ , because they are both equal to $90° - \theta$ .

For example, if $\theta = 30°$ , $\hat{B}$ and $\hat{E}$ would both be $60°$ . If $\theta = 50°$ , $\hat{B}$ and $\hat{E}$ would both be $40°$ . Since they both have the same algebraic expression, they will both have the same value no matter what value we substitute for $\theta$ .

The completed proof is:

In $\Delta ABC$ and $\Delta FEC$ :
1. $\angle C$ is common.
2. $\angle A = \angle F = 90°$ (because the lamp post and Eric are both perpendicular to the ground)
3. $\angle B = \angle E = 90° - \theta$ (sum of $\angle$ s in $\Delta$ )
$\therefore \Delta ABC \;|||\; \Delta FEC$ (equiangular $\Delta$ s)

Submit your answer as: andandand
We have been told that at $\text{8}\ \text{am}$ , Eric's shadow is $\text{2.65}\ \text{m}$ long and the lamp post's shadow is $\text{6.1}\ \text{m}$ long. We have also proved that $\Delta ABC \;|||\; \Delta FEC$ .

Hence, work out the height of the lamp post.

INSTRUCTION: Round your answer to two decimal places.

Answer:
Height of the lamp post $=$ $\text{m}$ $\;$ $(\Delta ABC \;|||\; \Delta FEC)$

one-of
type(numeric.abserror(0.01))
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Since the two triangles are similar, their sides are in proportion.

STEP: Use sides in proportion to create and solve an equation
[−3 points ⇒ 0 / 3 points left]

Since we have proved that $\Delta ABC \;|||\; \Delta FEC$ , their sides must be in proportion. We can use the ratio of sides to set up an equation.

Since the height of the lamp post ( $AB$ ) is what we want to find out, we will put it in the numerator. It matches with Eric's height ( $FE$ ) in the smaller triangle.

$\begin{aligned} \dfrac{AB}{\text{1.53}} & = \dfrac{\square}{\square} \end{aligned}$

In the same way, the lamp post's shadow ( $AC$ ) matches with Eric's shadow ( $FC$ ) in the smaller triangle.

$\begin{aligned} \dfrac{AB}{\text{1.53}} & = \dfrac{\text{6.1}}{\text{2.65}} \end{aligned}$

Hence,

$\begin{aligned} \dfrac{AB}{\text{1.53}} {\color{ForestGreen}{\, \times \, \text{1.53}}} & = \dfrac{\text{6.1}}{\text{2.65}} {\color{ForestGreen}{\, \times \, \text{1.53}}} \\ \therefore AB & = \text{3.52188...}\ \text{m} \\ & \approx \text{3.52}\ \text{m} \end{aligned}$

Submit your answer as:

ID is: 4314 Seed is: 359

Applications of similarity

Mike wants to find out the height of a tree. Mike is $\text{1.66}\ \text{m}$ tall. The length of Mike's shadow changes throughout the day.

At $\text{11}\ \text{am}$ , Mike stands in the shadow of the tree. The shadows of the tree and Mike form triangles. Both triangles contain the same angle, as shown in the diagram below.

At $\text{11}\ \text{am}$ , Mike's shadow is $\text{2.88}\ \text{m}$ long and the tree's shadow is $\text{6.91}\ \text{m}$ long. Complete the proof to help Mike work out the height of the tree.
Answer:
In $\Delta ABC$ and $\Delta FEC$ :
1. is common.
2. $\angle A = \angle CFE =$ $°$ (because the tree and Mike are both perpendicular to the ground)
3. $\angle B = \angle CEF =$ $°$ (sum of $\angle$ s in $\Delta$ )
$\therefore \Delta ABC \;|||\; \Delta FEC$

numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

It may help you to redraw the diagram with the two triangles separate from each other. Which pairs of angles are equal between the two triangles, and why?

STEP: Prove similarity
[−4 points ⇒ 0 / 4 points left]

TIP: In real-life questions like this one, we can always assume that vertical objects like trees and human beings are perpendicular to the ground. We also assume that the ground is perfectly horizontal!

Once we know that the angles at $B\hat{A}C$ and $E\hat{F}C$ are $90°$ , we can use sum of $\angle$ s in $\Delta$ to work out expressions for the angles at $A\hat{B}C$ and $F\hat{E}C$ .

In $\Delta ABC$ :

$\begin{aligned} {\color{MediumVioletRed}{\hat{A}}} + {\color{orange}{\hat{B}}} + {\color{teal}{\hat{C}}} & = 180° \; (\text{sum of } \angle \text{s in } \Delta) \\ \therefore 90° + \hat{B} + \theta & = 180° \\ \therefore 90° + \hat{B} + \theta {\color{blue}{-\; 90°}} {\color{red}{- \;\theta}} & = 180° {\color{blue}{-\; 90°}} {\color{red}{-\; \theta}} \\ \therefore \hat{B} & = 90° - \theta \end{aligned}$

Similarly, in $\Delta FEC$ :

$\begin{aligned} {\color{MediumVioletRed}{\hat{F}}} + {\color{orange}{\hat{E}}} + {\color{teal}{\hat{C}}} & = 180° \; (\text{sum of } \angle \text{s in } \Delta) \\ \therefore 90° + \hat{E} + \theta & = 180° \\ \therefore \hat{E} & = 90° - \theta \end{aligned}$

So, even though we don't know what $\theta$ is, we can say for certain that $\hat{B} = \hat{E}$ , because they are both equal to $90° - \theta$ .

For example, if $\theta = 30°$ , $\hat{B}$ and $\hat{E}$ would both be $60°$ . If $\theta = 50°$ , $\hat{B}$ and $\hat{E}$ would both be $40°$ . Since they both have the same algebraic expression, they will both have the same value no matter what value we substitute for $\theta$ .

The completed proof is:

In $\Delta ABC$ and $\Delta FEC$ :
1. $\angle C$ is common.
2. $\angle A = \angle F = 90°$ (because the tree and Mike are both perpendicular to the ground)
3. $\angle B = \angle E = 90° - \theta$ (sum of $\angle$ s in $\Delta$ )
$\therefore \Delta ABC \;|||\; \Delta FEC$ (equiangular $\Delta$ s)

Submit your answer as: andandand
We have been told that at $\text{11}\ \text{am}$ , Mike's shadow is $\text{2.88}\ \text{m}$ long and the tree's shadow is $\text{6.91}\ \text{m}$ long. We have also proved that $\Delta ABC \;|||\; \Delta FEC$ .

Hence, work out the height of the tree.

INSTRUCTION: Round your answer to two decimal places.

Answer:
Height of the tree $=$ $\text{m}$ $\;$ $(\Delta ABC \;|||\; \Delta FEC)$

one-of
type(numeric.abserror(0.01))
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Since the two triangles are similar, their sides are in proportion.

STEP: Use sides in proportion to create and solve an equation
[−3 points ⇒ 0 / 3 points left]

Since we have proved that $\Delta ABC \;|||\; \Delta FEC$ , their sides must be in proportion. We can use the ratio of sides to set up an equation.

Since the height of the tree ( $AB$ ) is what we want to find out, we will put it in the numerator. It matches with Mike's height ( $FE$ ) in the smaller triangle.

$\begin{aligned} \dfrac{AB}{\text{1.66}} & = \dfrac{\square}{\square} \end{aligned}$

In the same way, the tree's shadow ( $AC$ ) matches with Mike's shadow ( $FC$ ) in the smaller triangle.

$\begin{aligned} \dfrac{AB}{\text{1.66}} & = \dfrac{\text{6.91}}{\text{2.88}} \end{aligned}$

Hence,

$\begin{aligned} \dfrac{AB}{\text{1.66}} {\color{ForestGreen}{\, \times \, \text{1.66}}} & = \dfrac{\text{6.91}}{\text{2.88}} {\color{ForestGreen}{\, \times \, \text{1.66}}} \\ \therefore AB & = \text{3.98284...}\ \text{m} \\ & \approx \text{3.98}\ \text{m} \end{aligned}$

Submit your answer as: